Angles in a 3D ball of hexagons and squares (truncated octahedron)












1












$begingroup$


Trying to make a 3D-model of a "ball" made up of squares and hexagons:





Given a square and four hexagons connecting to the square sides, what is the angle required between the square plane and the hexagon planes to make two sides of each hexagon connect with the sides of the two neighbouring hexagons?






angle solid-geometry







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$endgroup$








  • 1




    $begingroup$
    The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 12:52










  • $begingroup$
    @Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 12:54








  • 1




    $begingroup$
    @JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:12






  • 1




    $begingroup$
    I agree, it is a truncated Octahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 13:22






  • 1




    $begingroup$
    The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:23
















1












$begingroup$


Trying to make a 3D-model of a "ball" made up of squares and hexagons:





Given a square and four hexagons connecting to the square sides, what is the angle required between the square plane and the hexagon planes to make two sides of each hexagon connect with the sides of the two neighbouring hexagons?






angle solid-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 12:52










  • $begingroup$
    @Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 12:54








  • 1




    $begingroup$
    @JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:12






  • 1




    $begingroup$
    I agree, it is a truncated Octahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 13:22






  • 1




    $begingroup$
    The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:23














1












1








1





$begingroup$


Trying to make a 3D-model of a "ball" made up of squares and hexagons:





Given a square and four hexagons connecting to the square sides, what is the angle required between the square plane and the hexagon planes to make two sides of each hexagon connect with the sides of the two neighbouring hexagons?






angle solid-geometry







share|cite|improve this question











$endgroup$




Trying to make a 3D-model of a "ball" made up of squares and hexagons:





Given a square and four hexagons connecting to the square sides, what is the angle required between the square plane and the hexagon planes to make two sides of each hexagon connect with the sides of the two neighbouring hexagons?






angle solid-geometry




angle solid-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 19 '17 at 10:07









grg

1,0431812




1,0431812










asked Apr 29 '17 at 12:39









DHaDHa

1084




1084








  • 1




    $begingroup$
    The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 12:52










  • $begingroup$
    @Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 12:54








  • 1




    $begingroup$
    @JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:12






  • 1




    $begingroup$
    I agree, it is a truncated Octahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 13:22






  • 1




    $begingroup$
    The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:23














  • 1




    $begingroup$
    The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 12:52










  • $begingroup$
    @Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 12:54








  • 1




    $begingroup$
    @JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:12






  • 1




    $begingroup$
    I agree, it is a truncated Octahedron.
    $endgroup$
    – Jean Marie
    Apr 29 '17 at 13:22






  • 1




    $begingroup$
    The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
    $endgroup$
    – Donald Splutterwit
    Apr 29 '17 at 13:23








1




1




$begingroup$
The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 12:52




$begingroup$
The dihedral angle of a cuboctahedron is $theta= cos^{-1} (frac{-1}{sqrt{3}}) simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 12:52












$begingroup$
@Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
$endgroup$
– Jean Marie
Apr 29 '17 at 12:54






$begingroup$
@Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron.
$endgroup$
– Jean Marie
Apr 29 '17 at 12:54






1




1




$begingroup$
@JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 13:12




$begingroup$
@JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $ddot smile$
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 13:12




1




1




$begingroup$
I agree, it is a truncated Octahedron.
$endgroup$
– Jean Marie
Apr 29 '17 at 13:22




$begingroup$
I agree, it is a truncated Octahedron.
$endgroup$
– Jean Marie
Apr 29 '17 at 13:22




1




1




$begingroup$
The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 13:23




$begingroup$
The angle between the Hexagonal faces is $phi= cos^{-1} (frac{-1}{3}) simeq 109.47$
$endgroup$
– Donald Splutterwit
Apr 29 '17 at 13:23










1 Answer
1






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oldest

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1












$begingroup$

As already stated in the comment, you're looking at part of a truncated octahedron





The angles between the faces are stated as well:




4-6: arccos(−1/√3) = 125°15′51″

6-6: arccos(−1/3) = 109°28′16″




If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,pm1,pm2)$ (note this gives you edges of length $sqrt2$).






share|cite|improve this answer











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    $begingroup$

    As already stated in the comment, you're looking at part of a truncated octahedron





    The angles between the faces are stated as well:




    4-6: arccos(−1/√3) = 125°15′51″

    6-6: arccos(−1/3) = 109°28′16″




    If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,pm1,pm2)$ (note this gives you edges of length $sqrt2$).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      As already stated in the comment, you're looking at part of a truncated octahedron





      The angles between the faces are stated as well:




      4-6: arccos(−1/√3) = 125°15′51″

      6-6: arccos(−1/3) = 109°28′16″




      If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,pm1,pm2)$ (note this gives you edges of length $sqrt2$).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        As already stated in the comment, you're looking at part of a truncated octahedron





        The angles between the faces are stated as well:




        4-6: arccos(−1/√3) = 125°15′51″

        6-6: arccos(−1/3) = 109°28′16″




        If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,pm1,pm2)$ (note this gives you edges of length $sqrt2$).






        share|cite|improve this answer











        $endgroup$



        As already stated in the comment, you're looking at part of a truncated octahedron





        The angles between the faces are stated as well:




        4-6: arccos(−1/√3) = 125°15′51″

        6-6: arccos(−1/3) = 109°28′16″




        If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,pm1,pm2)$ (note this gives you edges of length $sqrt2$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 '18 at 20:03

























        answered Dec 9 '18 at 18:33









        Tobias KienzlerTobias Kienzler

        3,4602662




        3,4602662






























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