What fraction of the points in a 24-cell of radius 1 are within 0.85 units of the center?

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Consider a 24-cell (one of the six regular polytopes in four dimensions) of radius 1. I want a formula for the fraction of all points inside the 24-cell that are within a certain distance $r$ from the center. I already worked out two formulas.



For $0 le r le sqrtfrac12$,the set of points is simply a 3-sphere inside the 24-cell. So the fraction formula is simply the 4-volume of the 3-sphere divided by the 4-volume of the 24-cell, which is $frac14pi^2r^4$.



For $sqrtfrac12 le r le sqrtfrac23$, the 3-sphere crosses outside the 24-cell, and it does so in such a way that we have 24 "domes" outside the polytope that we subtract off. I did the math and the formula for this is:



$frac14pi^2r^4 - 3piarccos(frac1r^2 - 1)r^4 + pi(5r^2 - 1)sqrt2r^2 - 1$



There are two more formulas for the ranges $[sqrtfrac23, sqrtfrac34]$ and $[sqrtfrac34, 1]$. Can anybody figure them out, at least one of them?




geometry




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    Consider a 24-cell (one of the six regular polytopes in four dimensions) of radius 1. I want a formula for the fraction of all points inside the 24-cell that are within a certain distance $r$ from the center. I already worked out two formulas.



    For $0 le r le sqrtfrac12$,the set of points is simply a 3-sphere inside the 24-cell. So the fraction formula is simply the 4-volume of the 3-sphere divided by the 4-volume of the 24-cell, which is $frac14pi^2r^4$.



    For $sqrtfrac12 le r le sqrtfrac23$, the 3-sphere crosses outside the 24-cell, and it does so in such a way that we have 24 "domes" outside the polytope that we subtract off. I did the math and the formula for this is:



    $frac14pi^2r^4 - 3piarccos(frac1r^2 - 1)r^4 + pi(5r^2 - 1)sqrt2r^2 - 1$



    There are two more formulas for the ranges $[sqrtfrac23, sqrtfrac34]$ and $[sqrtfrac34, 1]$. Can anybody figure them out, at least one of them?




    geometry




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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Consider a 24-cell (one of the six regular polytopes in four dimensions) of radius 1. I want a formula for the fraction of all points inside the 24-cell that are within a certain distance $r$ from the center. I already worked out two formulas.



      For $0 le r le sqrtfrac12$,the set of points is simply a 3-sphere inside the 24-cell. So the fraction formula is simply the 4-volume of the 3-sphere divided by the 4-volume of the 24-cell, which is $frac14pi^2r^4$.



      For $sqrtfrac12 le r le sqrtfrac23$, the 3-sphere crosses outside the 24-cell, and it does so in such a way that we have 24 "domes" outside the polytope that we subtract off. I did the math and the formula for this is:



      $frac14pi^2r^4 - 3piarccos(frac1r^2 - 1)r^4 + pi(5r^2 - 1)sqrt2r^2 - 1$



      There are two more formulas for the ranges $[sqrtfrac23, sqrtfrac34]$ and $[sqrtfrac34, 1]$. Can anybody figure them out, at least one of them?




      geometry




      share|cite|improve this question











      Consider a 24-cell (one of the six regular polytopes in four dimensions) of radius 1. I want a formula for the fraction of all points inside the 24-cell that are within a certain distance $r$ from the center. I already worked out two formulas.



      For $0 le r le sqrtfrac12$,the set of points is simply a 3-sphere inside the 24-cell. So the fraction formula is simply the 4-volume of the 3-sphere divided by the 4-volume of the 24-cell, which is $frac14pi^2r^4$.



      For $sqrtfrac12 le r le sqrtfrac23$, the 3-sphere crosses outside the 24-cell, and it does so in such a way that we have 24 "domes" outside the polytope that we subtract off. I did the math and the formula for this is:



      $frac14pi^2r^4 - 3piarccos(frac1r^2 - 1)r^4 + pi(5r^2 - 1)sqrt2r^2 - 1$



      There are two more formulas for the ranges $[sqrtfrac23, sqrtfrac34]$ and $[sqrtfrac34, 1]$. Can anybody figure them out, at least one of them?





      geometry





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      asked Jul 21 at 22:03









      Perry Ainsworth

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