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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













6












$begingroup$


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











$endgroup$








  • 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
















6












$begingroup$


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











$endgroup$








  • 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














6












6








6


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











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Ahmed Ashour

972313




972313










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














  • 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










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











$endgroup$





















    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











    $endgroup$





















      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











      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        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











        $endgroup$


















          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











          $endgroup$
















            5












            5








            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











            $endgroup$




            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








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 7 hours ago

























            answered 8 hours ago









            Weather VaneWeather Vane

            1,49919




            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











                $endgroup$


















                  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











                  $endgroup$
















                    5












                    5








                    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











                    $endgroup$



                    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.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 2 hours ago









                    2012rcampion

                    11.4k14273




                    11.4k14273










                    answered 9 hours ago









                    Jaap ScherphuisJaap Scherphuis

                    16.1k12772




                    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











                        $endgroup$


















                          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











                          $endgroup$
















                            4












                            4








                            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











                            $endgroup$



                            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) = $256$.








                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 3 hours ago









                            Hugh

                            2,26811127




                            2,26811127










                            answered 9 hours ago









                            Jan IvanJan Ivan

                            2,066620




                            2,066620






























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                                Castillo d'Acher Características Menú de navegación

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                                Montañas de los Pirineos de AragónMontañas de la provincia de Huesca msnmSelva de OzaPirineo aragonésEspañabosque Este artículo o sección necesita referencias que aparezcan en una publicación acreditada. Este aviso fue puesto el 11 de agosto de 2011. El pico desde Selva de Oza. El Castillo d' Acher es una cima de 2.384msnm en la zona de la Selva de Oza en la parte occidental del Pirineo aragonés, en España. Debe su nombre a su aspecto de fortaleza que presenta hacia todas sus vertientes. Características En orden ascendente, la cima pasa de un bosque frondoso de tonalidades verde-amarillenta a laderas rojizas. Su cima, que suele presentarse mucha nieve, es un pequeño valle rectangular rodeado en sus cuatro caras por una muralla que le da un aspecto de fortaleza. This page is only for reference, If you need detailed information, please check here
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