Calculate “Volume” and “Surface Area” of a voxel-based sphere
$begingroup$
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):
Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:
- The number of voxels in the sphere if it were hollow (kind of a 'surface area')
- The number of voxels in the sphere if it were solid (kind of a 'volume')
Is there a formula possible here? :)
geometry
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
$endgroup$
add a comment |
$begingroup$
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):
Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:
- The number of voxels in the sphere if it were hollow (kind of a 'surface area')
- The number of voxels in the sphere if it were solid (kind of a 'volume')
Is there a formula possible here? :)
geometry
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
$endgroup$
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
add a comment |
$begingroup$
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):
Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:
- The number of voxels in the sphere if it were hollow (kind of a 'surface area')
- The number of voxels in the sphere if it were solid (kind of a 'volume')
Is there a formula possible here? :)
geometry
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
$endgroup$
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):
Given the diameter of this voxel sphere (in number of voxels, e.g. '20 voxels in diameter'), how can I calculate:
- The number of voxels in the sphere if it were hollow (kind of a 'surface area')
- The number of voxels in the sphere if it were solid (kind of a 'volume')
Is there a formula possible here? :)
geometry
geometry
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
asked Nov 10 '11 at 2:56
InfinitiesLoopInfinitiesLoop
12614
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
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
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
$endgroup$
add a comment |
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$begingroup$
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
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
$endgroup$
add a comment |
$begingroup$
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
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
$endgroup$
add a comment |
$begingroup$
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
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
$endgroup$
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
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
answered Oct 5 '18 at 15:40
com.prehensiblecom.prehensible
18511
18511
add a comment |
add a comment |
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$begingroup$
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