Kelvin problem: Single polyhedron, non-isohedral [on hold]











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What is the solution to the Kelvin problem under the restriction of only one type of (not necessarily isohedral) polyhedron? Is it the truncated octahedron? Proof?






geometry polyhedra minimal-surfaces tessellations







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put on hold as off-topic by heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica 22 hours ago


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    up vote
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    What is the solution to the Kelvin problem under the restriction of only one type of (not necessarily isohedral) polyhedron? Is it the truncated octahedron? Proof?






    geometry polyhedra minimal-surfaces tessellations







    share|cite|improve this question













    put on hold as off-topic by heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica 22 hours ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica

    If this question can be reworded to fit the rules in the help center, please edit the question.















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      What is the solution to the Kelvin problem under the restriction of only one type of (not necessarily isohedral) polyhedron? Is it the truncated octahedron? Proof?






      geometry polyhedra minimal-surfaces tessellations







      share|cite|improve this question













      What is the solution to the Kelvin problem under the restriction of only one type of (not necessarily isohedral) polyhedron? Is it the truncated octahedron? Proof?






      geometry polyhedra minimal-surfaces tessellations




      geometry polyhedra minimal-surfaces tessellations






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      asked yesterday









      MrFrety

      8410




      8410




      put on hold as off-topic by heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica 22 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica 22 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, user10354138, Davide Giraudo, Paul Frost, Scientifica

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          The truncated octahedron is indeed the correct answer as there are only 5 types of parallelohedra, among which it is optimal.



          Cp. https://en.m.wikipedia.org/wiki/Parallelohedron






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote













            The truncated octahedron is indeed the correct answer as there are only 5 types of parallelohedra, among which it is optimal.



            Cp. https://en.m.wikipedia.org/wiki/Parallelohedron






            share|cite|improve this answer

























              up vote
              0
              down vote













              The truncated octahedron is indeed the correct answer as there are only 5 types of parallelohedra, among which it is optimal.



              Cp. https://en.m.wikipedia.org/wiki/Parallelohedron






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                The truncated octahedron is indeed the correct answer as there are only 5 types of parallelohedra, among which it is optimal.



                Cp. https://en.m.wikipedia.org/wiki/Parallelohedron






                share|cite|improve this answer












                The truncated octahedron is indeed the correct answer as there are only 5 types of parallelohedra, among which it is optimal.



                Cp. https://en.m.wikipedia.org/wiki/Parallelohedron







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                MrFrety

                8410




                8410















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