Calculate “Volume” and “Surface Area” of a voxel-based sphere












5












$begingroup$


By "voxel-based sphere" I mean a sphere made up of cubes. Sorry if that is not the correct terminology. Imagine a sphere made out of legos. Except each voxel is a cube (unlike most legos). Determining the voxel distribution to make the sphere I suppose would involve calculating the x/y/z of a position on the sphere and then 'snapping' it to the nearest multiple of the voxel width/height.



Here's a calculator that can generate one:
http://neil.fraser.name/news/2006/11/17/



And here is an image of one (cross sectioned):



enter image description here



Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:




  1. The number of voxels in the sphere if it were hollow (kind of a 'surface area')

  2. The number of voxels in the sphere if it were solid (kind of a 'volume')


Is there a formula possible here? :)






geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
    $endgroup$
    – anon
    Nov 10 '11 at 3:03
















5












$begingroup$


By "voxel-based sphere" I mean a sphere made up of cubes. Sorry if that is not the correct terminology. Imagine a sphere made out of legos. Except each voxel is a cube (unlike most legos). Determining the voxel distribution to make the sphere I suppose would involve calculating the x/y/z of a position on the sphere and then 'snapping' it to the nearest multiple of the voxel width/height.



Here's a calculator that can generate one:
http://neil.fraser.name/news/2006/11/17/



And here is an image of one (cross sectioned):



enter image description here



Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:




  1. The number of voxels in the sphere if it were hollow (kind of a 'surface area')

  2. The number of voxels in the sphere if it were solid (kind of a 'volume')


Is there a formula possible here? :)






geometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
    $endgroup$
    – anon
    Nov 10 '11 at 3:03














5












5








5





$begingroup$


By "voxel-based sphere" I mean a sphere made up of cubes. Sorry if that is not the correct terminology. Imagine a sphere made out of legos. Except each voxel is a cube (unlike most legos). Determining the voxel distribution to make the sphere I suppose would involve calculating the x/y/z of a position on the sphere and then 'snapping' it to the nearest multiple of the voxel width/height.



Here's a calculator that can generate one:
http://neil.fraser.name/news/2006/11/17/



And here is an image of one (cross sectioned):



enter image description here



Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:




  1. The number of voxels in the sphere if it were hollow (kind of a 'surface area')

  2. The number of voxels in the sphere if it were solid (kind of a 'volume')


Is there a formula possible here? :)






geometry







share|cite|improve this question









$endgroup$




By "voxel-based sphere" I mean a sphere made up of cubes. Sorry if that is not the correct terminology. Imagine a sphere made out of legos. Except each voxel is a cube (unlike most legos). Determining the voxel distribution to make the sphere I suppose would involve calculating the x/y/z of a position on the sphere and then 'snapping' it to the nearest multiple of the voxel width/height.



Here's a calculator that can generate one:
http://neil.fraser.name/news/2006/11/17/



And here is an image of one (cross sectioned):



enter image description here



Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:




  1. The number of voxels in the sphere if it were hollow (kind of a 'surface area')

  2. The number of voxels in the sphere if it were solid (kind of a 'volume')


Is there a formula possible here? :)






geometry




geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 10 '11 at 2:56









InfinitiesLoopInfinitiesLoop

12614




12614








  • 1




    $begingroup$
    Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
    $endgroup$
    – anon
    Nov 10 '11 at 3:03














  • 1




    $begingroup$
    Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
    $endgroup$
    – anon
    Nov 10 '11 at 3:03








1




1




$begingroup$
Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
$endgroup$
– anon
Nov 10 '11 at 3:03




$begingroup$
Decompose it into cross sections; each segment should allow an explicit summation involving square roots and floor/ceiling functions.
$endgroup$
– anon
Nov 10 '11 at 3:03










1 Answer
1






active

oldest

votes


















0












$begingroup$

That's very easy, it's a function of an XYZ loop....



You would be actually doing a marching cubes kind of logic, and an ISOsurface logic for a spherical volume.



The first result is this programming query (in metacode):



var cubecount:
ForLoop(x,y,z){
If space(x,y,z) corresponds to DistanceFunction(x,y,z) == 1;
Then cubecount +=1:



question 2 is the same using >=1



For a sphere on zero of radius 1. (distancefunction is xposvectormagnitude, same thing if the sphere is origin is 0; otherwise the sphere pos equation is for the xyz loop is:



This code makes a 3d printable off axis voxel ball if you want:



num = 45.9874;



for (x =[-num-5:num+5]) 

{

    for (y =[-num-5:num+5]) 

    {    

        for (z =[-num-5:num+5]) 

        {    

              if ((x+1.24)*(x+1.24)+(y+1.66)*(y+1.66)+(z+1.88)*(z+1.88)<=num)  

                translate([x,y,z])    

                cube(1);





        }





    }

}


https://www.thingiverse.com/thing:3075040






share|cite|improve this answer









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    1 Answer
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    active

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    0












    $begingroup$

    That's very easy, it's a function of an XYZ loop....



    You would be actually doing a marching cubes kind of logic, and an ISOsurface logic for a spherical volume.



    The first result is this programming query (in metacode):



    var cubecount:
    ForLoop(x,y,z){
    If space(x,y,z) corresponds to DistanceFunction(x,y,z) == 1;
    Then cubecount +=1:



    question 2 is the same using >=1



    For a sphere on zero of radius 1. (distancefunction is xposvectormagnitude, same thing if the sphere is origin is 0; otherwise the sphere pos equation is for the xyz loop is:



    This code makes a 3d printable off axis voxel ball if you want:



    num = 45.9874;
    

    
for (x =[-num-5:num+5]) 
    
{
    
    for (y =[-num-5:num+5]) 
    
    {    
    
        for (z =[-num-5:num+5]) 
    
        {    
    
              if ((x+1.24)*(x+1.24)+(y+1.66)*(y+1.66)+(z+1.88)*(z+1.88)<=num)  
    
                translate([x,y,z])    
    
                cube(1);
    

    

    
        }
    

    

    
    }
    
}


    https://www.thingiverse.com/thing:3075040






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      That's very easy, it's a function of an XYZ loop....



      You would be actually doing a marching cubes kind of logic, and an ISOsurface logic for a spherical volume.



      The first result is this programming query (in metacode):



      var cubecount:
      ForLoop(x,y,z){
      If space(x,y,z) corresponds to DistanceFunction(x,y,z) == 1;
      Then cubecount +=1:



      question 2 is the same using >=1



      For a sphere on zero of radius 1. (distancefunction is xposvectormagnitude, same thing if the sphere is origin is 0; otherwise the sphere pos equation is for the xyz loop is:



      This code makes a 3d printable off axis voxel ball if you want:



      num = 45.9874;
      

      
for (x =[-num-5:num+5]) 
      
{
      
    for (y =[-num-5:num+5]) 
      
    {    
      
        for (z =[-num-5:num+5]) 
      
        {    
      
              if ((x+1.24)*(x+1.24)+(y+1.66)*(y+1.66)+(z+1.88)*(z+1.88)<=num)  
      
                translate([x,y,z])    
      
                cube(1);
      

      

      
        }
      

      

      
    }
      
}


      https://www.thingiverse.com/thing:3075040






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        That's very easy, it's a function of an XYZ loop....



        You would be actually doing a marching cubes kind of logic, and an ISOsurface logic for a spherical volume.



        The first result is this programming query (in metacode):



        var cubecount:
        ForLoop(x,y,z){
        If space(x,y,z) corresponds to DistanceFunction(x,y,z) == 1;
        Then cubecount +=1:



        question 2 is the same using >=1



        For a sphere on zero of radius 1. (distancefunction is xposvectormagnitude, same thing if the sphere is origin is 0; otherwise the sphere pos equation is for the xyz loop is:



        This code makes a 3d printable off axis voxel ball if you want:



        num = 45.9874;
        

        
for (x =[-num-5:num+5]) 
        
{
        
    for (y =[-num-5:num+5]) 
        
    {    
        
        for (z =[-num-5:num+5]) 
        
        {    
        
              if ((x+1.24)*(x+1.24)+(y+1.66)*(y+1.66)+(z+1.88)*(z+1.88)<=num)  
        
                translate([x,y,z])    
        
                cube(1);
        

        

        
        }
        

        

        
    }
        
}


        https://www.thingiverse.com/thing:3075040






        share|cite|improve this answer









        $endgroup$



        That's very easy, it's a function of an XYZ loop....



        You would be actually doing a marching cubes kind of logic, and an ISOsurface logic for a spherical volume.



        The first result is this programming query (in metacode):



        var cubecount:
        ForLoop(x,y,z){
        If space(x,y,z) corresponds to DistanceFunction(x,y,z) == 1;
        Then cubecount +=1:



        question 2 is the same using >=1



        For a sphere on zero of radius 1. (distancefunction is xposvectormagnitude, same thing if the sphere is origin is 0; otherwise the sphere pos equation is for the xyz loop is:



        This code makes a 3d printable off axis voxel ball if you want:



        num = 45.9874;
        

        
for (x =[-num-5:num+5]) 
        
{
        
    for (y =[-num-5:num+5]) 
        
    {    
        
        for (z =[-num-5:num+5]) 
        
        {    
        
              if ((x+1.24)*(x+1.24)+(y+1.66)*(y+1.66)+(z+1.88)*(z+1.88)<=num)  
        
                translate([x,y,z])    
        
                cube(1);
        

        

        
        }
        

        

        
    }
        
}


        https://www.thingiverse.com/thing:3075040







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 5 '18 at 15:40









        com.prehensiblecom.prehensible

        18511




        18511






























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