Equal-area sparse spherical shell partitioning
$begingroup$
I'm trying to solve a particular problem that arose in a computer graphics context, but can be generalised to a bigger problem as well. I'm not entirely sure if this question belongs to MathExchange either, so any suggestions are greatly appreciated.
If you with to skip the preamble, fell free to go to the "Main Problem" section.
The original problem
(a little background, for the ones not inclined to computer graphics: bump-maps are simply computer images in which each pixel stores a normalised three-dimensional vector, instead of three discrete color intensity values)
I got an idea the other day, when trying to optimize bump/normal mapping on a very computing-power-restricted environment, to replace my RGB bump-map with an indexed image. This way, instead of calculating the lightning equation/dot product for each pixel on display, I would just need to calculate it for each normal stored on the image index, and replace the image index by the equivalent light intensity obtained for each normal when displaying the image. It has obvious shortcomings, such as the need for an infinitely distant (or parallel) light source (instead of point light sources), but for my purposes it works just fine.
So, in a new attempt to reduce even further the computing power required for the operation, and compress the map even further, I imagined what would be needed to generate an ideal index, given an space constraint (amount of stored normals). Turns out it's the very key idea about my question that will solve this as well.
The main problem
The above problem can be solved for N indexes if there's a way to partition a spherical shell of unitary radius into N equal-area shell segments/cuts, much like an optimised Voronoi Diagram, albeit in spherical space. A turtle shell is probably the best real-world analog to this idea:
My question basically boils down to, "Is there a way to partition a spherical shell into N polygonal segments, akin to a Voronoi Diagram (where the edges of the polygons themselves are straight arcs on the shell surface and the regions themselves are as "uniform" and sparse as possible on the shell surface), where each of those segments occupy exactly the same shell area?"
Considerations
- I assume a solution to this problem would be dependent on some sort of constraint, like a set of initial vectors or orientations. If possible, I'm asking for the method which requires the least amount of user-supplied data.
- Randomized methods are not an option; I'm looking for an exact and repeatable solution.
vector-spaces spherical-geometry
$endgroup$
add a comment |
$begingroup$
I'm trying to solve a particular problem that arose in a computer graphics context, but can be generalised to a bigger problem as well. I'm not entirely sure if this question belongs to MathExchange either, so any suggestions are greatly appreciated.
If you with to skip the preamble, fell free to go to the "Main Problem" section.
The original problem
(a little background, for the ones not inclined to computer graphics: bump-maps are simply computer images in which each pixel stores a normalised three-dimensional vector, instead of three discrete color intensity values)
I got an idea the other day, when trying to optimize bump/normal mapping on a very computing-power-restricted environment, to replace my RGB bump-map with an indexed image. This way, instead of calculating the lightning equation/dot product for each pixel on display, I would just need to calculate it for each normal stored on the image index, and replace the image index by the equivalent light intensity obtained for each normal when displaying the image. It has obvious shortcomings, such as the need for an infinitely distant (or parallel) light source (instead of point light sources), but for my purposes it works just fine.
So, in a new attempt to reduce even further the computing power required for the operation, and compress the map even further, I imagined what would be needed to generate an ideal index, given an space constraint (amount of stored normals). Turns out it's the very key idea about my question that will solve this as well.
The main problem
The above problem can be solved for N indexes if there's a way to partition a spherical shell of unitary radius into N equal-area shell segments/cuts, much like an optimised Voronoi Diagram, albeit in spherical space. A turtle shell is probably the best real-world analog to this idea:
My question basically boils down to, "Is there a way to partition a spherical shell into N polygonal segments, akin to a Voronoi Diagram (where the edges of the polygons themselves are straight arcs on the shell surface and the regions themselves are as "uniform" and sparse as possible on the shell surface), where each of those segments occupy exactly the same shell area?"
Considerations
- I assume a solution to this problem would be dependent on some sort of constraint, like a set of initial vectors or orientations. If possible, I'm asking for the method which requires the least amount of user-supplied data.
- Randomized methods are not an option; I'm looking for an exact and repeatable solution.
vector-spaces spherical-geometry
$endgroup$
add a comment |
$begingroup$
I'm trying to solve a particular problem that arose in a computer graphics context, but can be generalised to a bigger problem as well. I'm not entirely sure if this question belongs to MathExchange either, so any suggestions are greatly appreciated.
If you with to skip the preamble, fell free to go to the "Main Problem" section.
The original problem
(a little background, for the ones not inclined to computer graphics: bump-maps are simply computer images in which each pixel stores a normalised three-dimensional vector, instead of three discrete color intensity values)
I got an idea the other day, when trying to optimize bump/normal mapping on a very computing-power-restricted environment, to replace my RGB bump-map with an indexed image. This way, instead of calculating the lightning equation/dot product for each pixel on display, I would just need to calculate it for each normal stored on the image index, and replace the image index by the equivalent light intensity obtained for each normal when displaying the image. It has obvious shortcomings, such as the need for an infinitely distant (or parallel) light source (instead of point light sources), but for my purposes it works just fine.
So, in a new attempt to reduce even further the computing power required for the operation, and compress the map even further, I imagined what would be needed to generate an ideal index, given an space constraint (amount of stored normals). Turns out it's the very key idea about my question that will solve this as well.
The main problem
The above problem can be solved for N indexes if there's a way to partition a spherical shell of unitary radius into N equal-area shell segments/cuts, much like an optimised Voronoi Diagram, albeit in spherical space. A turtle shell is probably the best real-world analog to this idea:
My question basically boils down to, "Is there a way to partition a spherical shell into N polygonal segments, akin to a Voronoi Diagram (where the edges of the polygons themselves are straight arcs on the shell surface and the regions themselves are as "uniform" and sparse as possible on the shell surface), where each of those segments occupy exactly the same shell area?"
Considerations
- I assume a solution to this problem would be dependent on some sort of constraint, like a set of initial vectors or orientations. If possible, I'm asking for the method which requires the least amount of user-supplied data.
- Randomized methods are not an option; I'm looking for an exact and repeatable solution.
vector-spaces spherical-geometry
$endgroup$
I'm trying to solve a particular problem that arose in a computer graphics context, but can be generalised to a bigger problem as well. I'm not entirely sure if this question belongs to MathExchange either, so any suggestions are greatly appreciated.
If you with to skip the preamble, fell free to go to the "Main Problem" section.
The original problem
(a little background, for the ones not inclined to computer graphics: bump-maps are simply computer images in which each pixel stores a normalised three-dimensional vector, instead of three discrete color intensity values)
I got an idea the other day, when trying to optimize bump/normal mapping on a very computing-power-restricted environment, to replace my RGB bump-map with an indexed image. This way, instead of calculating the lightning equation/dot product for each pixel on display, I would just need to calculate it for each normal stored on the image index, and replace the image index by the equivalent light intensity obtained for each normal when displaying the image. It has obvious shortcomings, such as the need for an infinitely distant (or parallel) light source (instead of point light sources), but for my purposes it works just fine.
So, in a new attempt to reduce even further the computing power required for the operation, and compress the map even further, I imagined what would be needed to generate an ideal index, given an space constraint (amount of stored normals). Turns out it's the very key idea about my question that will solve this as well.
The main problem
The above problem can be solved for N indexes if there's a way to partition a spherical shell of unitary radius into N equal-area shell segments/cuts, much like an optimised Voronoi Diagram, albeit in spherical space. A turtle shell is probably the best real-world analog to this idea:
My question basically boils down to, "Is there a way to partition a spherical shell into N polygonal segments, akin to a Voronoi Diagram (where the edges of the polygons themselves are straight arcs on the shell surface and the regions themselves are as "uniform" and sparse as possible on the shell surface), where each of those segments occupy exactly the same shell area?"
Considerations
- I assume a solution to this problem would be dependent on some sort of constraint, like a set of initial vectors or orientations. If possible, I'm asking for the method which requires the least amount of user-supplied data.
- Randomized methods are not an option; I'm looking for an exact and repeatable solution.
vector-spaces spherical-geometry
vector-spaces spherical-geometry
edited Dec 24 '18 at 9:02
Glorfindel
3,41981830
3,41981830
asked Dec 28 '14 at 1:50
MVittiSMVittiS
1234
1234
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A similar problem has been studied before here: take a look at the last page of the article, where the partitions of the sphere are presented. The problem is to minimize the sum of polygon perimeters while keeping constant areas. The result is a hexagonal + $12$ pentagons packing (for $N geq 14$). I'm not sure how high you could get with $N$. In the paper they do it for $N leq 32$. For large values it can be hard to do such an optimization.
I did some work recently on this: http://www.lama.univ-savoie.fr/~bogosel/modica_mortola_sphere.html The same observation: it works until $N=32$, maybe even higher, but for arbitrarily large $N$ the computations get slow.
Another uniform partition of the sphere in equal areas is presented here:
http://fr.mathworks.com/matlabcentral/fileexchange/13356-eqsp--recursive-zonal-sphere-partitioning-toolbox This is fast, even for $N$ greater than a milion. The advantage is that it works with spherical rectangles, so each set of the partition is governed by only a few parameters.
$endgroup$
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
add a comment |
$begingroup$
This doesn't work for arbitrary $N$, but ...
Take an inscribed icosahedron, and project its edges outwards onto the surface of the sphere. This will give you 20 identical "equilateral" spherical triangles that cover the sphere. If you need more triangles, subdivide these 20. Division into 3 is easy, so you can get partionings that consist of 20, 60, 180, 540 pieces, and so on.
You could do the same sort of thing starting with a tetrahedron, actually. This would give you partions with 4, 12, 36, 108 pieces, and so on.
In fact, I guess you could use any Platonic solid as a starting point.
Subdividing triangles into 4 smaller ones (rather than 3) might be better, since the smaller triangles will then be more nearly equilateral.
$endgroup$
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
add a comment |
$begingroup$
If it's just about equal area pieces the following is simplest:
Let $N=pcdot q$. Divide the interval $[{-1},1]$ on the $z$-axis into $p$ equal parts $[z_{k-1},z_k]$ and intersect the sphere $S^2$ with the equidistant planes $z=z_k$. Then intersect $S^2$ with $q$ meridianal half planes at equal angles ${2piover q}$. This will produce $2q$ spherical triangles and $N-2q$ "spherical rectangles", all of the same area.
When you want "rounder" pieces, look up here:
http://en.wikipedia.org/wiki/Fullerene
Here the sphere is tessellated in a honeycomb manner, with twelve exceptional pentagons.
$endgroup$
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A similar problem has been studied before here: take a look at the last page of the article, where the partitions of the sphere are presented. The problem is to minimize the sum of polygon perimeters while keeping constant areas. The result is a hexagonal + $12$ pentagons packing (for $N geq 14$). I'm not sure how high you could get with $N$. In the paper they do it for $N leq 32$. For large values it can be hard to do such an optimization.
I did some work recently on this: http://www.lama.univ-savoie.fr/~bogosel/modica_mortola_sphere.html The same observation: it works until $N=32$, maybe even higher, but for arbitrarily large $N$ the computations get slow.
Another uniform partition of the sphere in equal areas is presented here:
http://fr.mathworks.com/matlabcentral/fileexchange/13356-eqsp--recursive-zonal-sphere-partitioning-toolbox This is fast, even for $N$ greater than a milion. The advantage is that it works with spherical rectangles, so each set of the partition is governed by only a few parameters.
$endgroup$
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
add a comment |
$begingroup$
A similar problem has been studied before here: take a look at the last page of the article, where the partitions of the sphere are presented. The problem is to minimize the sum of polygon perimeters while keeping constant areas. The result is a hexagonal + $12$ pentagons packing (for $N geq 14$). I'm not sure how high you could get with $N$. In the paper they do it for $N leq 32$. For large values it can be hard to do such an optimization.
I did some work recently on this: http://www.lama.univ-savoie.fr/~bogosel/modica_mortola_sphere.html The same observation: it works until $N=32$, maybe even higher, but for arbitrarily large $N$ the computations get slow.
Another uniform partition of the sphere in equal areas is presented here:
http://fr.mathworks.com/matlabcentral/fileexchange/13356-eqsp--recursive-zonal-sphere-partitioning-toolbox This is fast, even for $N$ greater than a milion. The advantage is that it works with spherical rectangles, so each set of the partition is governed by only a few parameters.
$endgroup$
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
add a comment |
$begingroup$
A similar problem has been studied before here: take a look at the last page of the article, where the partitions of the sphere are presented. The problem is to minimize the sum of polygon perimeters while keeping constant areas. The result is a hexagonal + $12$ pentagons packing (for $N geq 14$). I'm not sure how high you could get with $N$. In the paper they do it for $N leq 32$. For large values it can be hard to do such an optimization.
I did some work recently on this: http://www.lama.univ-savoie.fr/~bogosel/modica_mortola_sphere.html The same observation: it works until $N=32$, maybe even higher, but for arbitrarily large $N$ the computations get slow.
Another uniform partition of the sphere in equal areas is presented here:
http://fr.mathworks.com/matlabcentral/fileexchange/13356-eqsp--recursive-zonal-sphere-partitioning-toolbox This is fast, even for $N$ greater than a milion. The advantage is that it works with spherical rectangles, so each set of the partition is governed by only a few parameters.
$endgroup$
A similar problem has been studied before here: take a look at the last page of the article, where the partitions of the sphere are presented. The problem is to minimize the sum of polygon perimeters while keeping constant areas. The result is a hexagonal + $12$ pentagons packing (for $N geq 14$). I'm not sure how high you could get with $N$. In the paper they do it for $N leq 32$. For large values it can be hard to do such an optimization.
I did some work recently on this: http://www.lama.univ-savoie.fr/~bogosel/modica_mortola_sphere.html The same observation: it works until $N=32$, maybe even higher, but for arbitrarily large $N$ the computations get slow.
Another uniform partition of the sphere in equal areas is presented here:
http://fr.mathworks.com/matlabcentral/fileexchange/13356-eqsp--recursive-zonal-sphere-partitioning-toolbox This is fast, even for $N$ greater than a milion. The advantage is that it works with spherical rectangles, so each set of the partition is governed by only a few parameters.
edited Apr 10 '15 at 20:03
answered Apr 8 '15 at 21:53
Beni BogoselBeni Bogosel
17.5k346111
17.5k346111
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
add a comment |
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
$begingroup$
This is probably the closest possible answer to the original question! Thanks for the references, and good luck finishing your work/thesis on the subject.
$endgroup$
– MVittiS
Apr 14 '15 at 2:03
add a comment |
$begingroup$
This doesn't work for arbitrary $N$, but ...
Take an inscribed icosahedron, and project its edges outwards onto the surface of the sphere. This will give you 20 identical "equilateral" spherical triangles that cover the sphere. If you need more triangles, subdivide these 20. Division into 3 is easy, so you can get partionings that consist of 20, 60, 180, 540 pieces, and so on.
You could do the same sort of thing starting with a tetrahedron, actually. This would give you partions with 4, 12, 36, 108 pieces, and so on.
In fact, I guess you could use any Platonic solid as a starting point.
Subdividing triangles into 4 smaller ones (rather than 3) might be better, since the smaller triangles will then be more nearly equilateral.
$endgroup$
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
add a comment |
$begingroup$
This doesn't work for arbitrary $N$, but ...
Take an inscribed icosahedron, and project its edges outwards onto the surface of the sphere. This will give you 20 identical "equilateral" spherical triangles that cover the sphere. If you need more triangles, subdivide these 20. Division into 3 is easy, so you can get partionings that consist of 20, 60, 180, 540 pieces, and so on.
You could do the same sort of thing starting with a tetrahedron, actually. This would give you partions with 4, 12, 36, 108 pieces, and so on.
In fact, I guess you could use any Platonic solid as a starting point.
Subdividing triangles into 4 smaller ones (rather than 3) might be better, since the smaller triangles will then be more nearly equilateral.
$endgroup$
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
add a comment |
$begingroup$
This doesn't work for arbitrary $N$, but ...
Take an inscribed icosahedron, and project its edges outwards onto the surface of the sphere. This will give you 20 identical "equilateral" spherical triangles that cover the sphere. If you need more triangles, subdivide these 20. Division into 3 is easy, so you can get partionings that consist of 20, 60, 180, 540 pieces, and so on.
You could do the same sort of thing starting with a tetrahedron, actually. This would give you partions with 4, 12, 36, 108 pieces, and so on.
In fact, I guess you could use any Platonic solid as a starting point.
Subdividing triangles into 4 smaller ones (rather than 3) might be better, since the smaller triangles will then be more nearly equilateral.
$endgroup$
This doesn't work for arbitrary $N$, but ...
Take an inscribed icosahedron, and project its edges outwards onto the surface of the sphere. This will give you 20 identical "equilateral" spherical triangles that cover the sphere. If you need more triangles, subdivide these 20. Division into 3 is easy, so you can get partionings that consist of 20, 60, 180, 540 pieces, and so on.
You could do the same sort of thing starting with a tetrahedron, actually. This would give you partions with 4, 12, 36, 108 pieces, and so on.
In fact, I guess you could use any Platonic solid as a starting point.
Subdividing triangles into 4 smaller ones (rather than 3) might be better, since the smaller triangles will then be more nearly equilateral.
edited Dec 28 '14 at 12:40
answered Dec 28 '14 at 5:23
bubbabubba
30.4k33086
30.4k33086
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
add a comment |
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
$begingroup$
This was actually the first idea which came to my head, and is indeed a good way to approximate a "uniform" sphere (where the vertices are as equally distributed over the surface as possible). But, as you said, doesn't work for arbitrary N.
$endgroup$
– MVittiS
Dec 28 '14 at 11:54
add a comment |
$begingroup$
If it's just about equal area pieces the following is simplest:
Let $N=pcdot q$. Divide the interval $[{-1},1]$ on the $z$-axis into $p$ equal parts $[z_{k-1},z_k]$ and intersect the sphere $S^2$ with the equidistant planes $z=z_k$. Then intersect $S^2$ with $q$ meridianal half planes at equal angles ${2piover q}$. This will produce $2q$ spherical triangles and $N-2q$ "spherical rectangles", all of the same area.
When you want "rounder" pieces, look up here:
http://en.wikipedia.org/wiki/Fullerene
Here the sphere is tessellated in a honeycomb manner, with twelve exceptional pentagons.
$endgroup$
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
add a comment |
$begingroup$
If it's just about equal area pieces the following is simplest:
Let $N=pcdot q$. Divide the interval $[{-1},1]$ on the $z$-axis into $p$ equal parts $[z_{k-1},z_k]$ and intersect the sphere $S^2$ with the equidistant planes $z=z_k$. Then intersect $S^2$ with $q$ meridianal half planes at equal angles ${2piover q}$. This will produce $2q$ spherical triangles and $N-2q$ "spherical rectangles", all of the same area.
When you want "rounder" pieces, look up here:
http://en.wikipedia.org/wiki/Fullerene
Here the sphere is tessellated in a honeycomb manner, with twelve exceptional pentagons.
$endgroup$
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
add a comment |
$begingroup$
If it's just about equal area pieces the following is simplest:
Let $N=pcdot q$. Divide the interval $[{-1},1]$ on the $z$-axis into $p$ equal parts $[z_{k-1},z_k]$ and intersect the sphere $S^2$ with the equidistant planes $z=z_k$. Then intersect $S^2$ with $q$ meridianal half planes at equal angles ${2piover q}$. This will produce $2q$ spherical triangles and $N-2q$ "spherical rectangles", all of the same area.
When you want "rounder" pieces, look up here:
http://en.wikipedia.org/wiki/Fullerene
Here the sphere is tessellated in a honeycomb manner, with twelve exceptional pentagons.
$endgroup$
If it's just about equal area pieces the following is simplest:
Let $N=pcdot q$. Divide the interval $[{-1},1]$ on the $z$-axis into $p$ equal parts $[z_{k-1},z_k]$ and intersect the sphere $S^2$ with the equidistant planes $z=z_k$. Then intersect $S^2$ with $q$ meridianal half planes at equal angles ${2piover q}$. This will produce $2q$ spherical triangles and $N-2q$ "spherical rectangles", all of the same area.
When you want "rounder" pieces, look up here:
http://en.wikipedia.org/wiki/Fullerene
Here the sphere is tessellated in a honeycomb manner, with twelve exceptional pentagons.
edited Dec 29 '14 at 8:55
answered Dec 28 '14 at 13:03
Christian BlatterChristian Blatter
173k7113326
173k7113326
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
add a comment |
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
This is similar to the naïve sphere-approximation algorithm, if I'm not mistaken! The problem with this solution is that, while all the upper and lower spherical triangles indeed have the same area, the spherical rectangles do not; they even don't have the same area as the aforementioned triangles. Here you can see a sample sphere generated with your method; notice how the rectangles near the equator have "more area" overall than the ones closer to the poles.
$endgroup$
– MVittiS
Dec 28 '14 at 17:53
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
While the above approach is admittedly naive it certainly produces patches of equal area. It is a basic fact of spherical geometry that the area of spherical "lampshades" is proportional to their height $Delta z$.
$endgroup$
– Christian Blatter
Dec 28 '14 at 19:33
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
My apologies; I've spent at least an hour convincing myself that your answer is, technically, correct. However, it still doesn't come close to the answer I was expecting, and which is a fault in the phrasing of my main question: I was hoping to obtain a sparsest and most uniform possible polygonal distribution, which is why I cited both the Voronoi Diagrams and the Turtle Shells. This is my first question on MathExchange, in reality, and now I don't know if it would be better to mark your answer as the correct one and make a new question, or to edit this one.
$endgroup$
– MVittiS
Dec 28 '14 at 23:41
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
$begingroup$
The Buckyball is certainly a step towards my desired outcome; if there was a way to generalize a Buckyball to a $N$-sided structure, then we have a perfectly reasonable answer to my question.
$endgroup$
– MVittiS
Dec 29 '14 at 22:10
add a comment |
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