A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges. Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ? (A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
="Henry Baker" <hbaker1@pipeline.com> A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges.
Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ?
(A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
How would we quadrangulate a tetrahedron?
I was thinking more along the lines of approximating smooth surfaces. I guess one could break up the triangles on the surface of the tetrahedron into quadrilaterals, but that method would enable any triangulated surface to be broken up in the same way. I was hoping for something less trivial. At 08:53 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges.
Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ?
(A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
How would we quadrangulate a tetrahedron?
Let's add another rule: quadrilaterals that meet at a zero angle at their edges become a larger planar polygon; edges that meet at a zero angle become a longer edge. So a tetrahedron can't be quadrangulated. Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface? The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation? At 09:00 AM 1/6/2013, Henry Baker wrote:
I was thinking more along the lines of approximating smooth surfaces.
I guess one could break up the triangles on the surface of the tetrahedron into quadrilaterals, but that method would enable any triangulated surface to be broken up in the same way.
I was hoping for something less trivial.
At 08:53 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges.
Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ?
(A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
How would we quadrangulate a tetrahedron?
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-). Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be. And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
Good, but we have a convex surface modelled by a non-convex surface. Can we do better? Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals. At 10:31 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
Erm - I think a better solution would be split the initial quadrilateral into thirds - this also allows correction of the initial scale edges, --------- | | | | --------- | | | | --------- | | | | --------- On 6 Jan 2013, at 18:39, Henry Baker wrote:
Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
At 10:31 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
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Hmmm... Are you sure that when you "push out" the middle square, that all of the quadrilaterals remain planar? At 10:47 AM 1/6/2013, David Makin wrote:
Erm - I think a better solution would be split the initial quadrilateral into thirds - this also allows correction of the initial scale edges,
--------- | | | | --------- | | | | --------- | | | | ---------
On 6 Jan 2013, at 18:39, Henry Baker wrote:
Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
At 10:31 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
No, you'd have to force that by approximation - but then again they don't in the previous example unless the smaller rectangle is coplanar with the original (and if not coplanar then I don't think you could even force an approximation). On 6 Jan 2013, at 18:50, Henry Baker wrote:
Hmmm... Are you sure that when you "push out" the middle square, that all of the quadrilaterals remain planar?
At 10:47 AM 1/6/2013, David Makin wrote:
Erm - I think a better solution would be split the initial quadrilateral into thirds - this also allows correction of the initial scale edges,
--------- | | | | --------- | | | | --------- | | | | ---------
On 6 Jan 2013, at 18:39, Henry Baker wrote:
Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
At 10:31 AM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
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="Henry Baker" <hbaker1@pipeline.com> Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Well, how do we define "better"? On the one hand we seem to be talking about approximating surfaces ever more closely, but on the other we seem to be groping to converge on discovering some unstated intuitive definition.
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
Then you just get the dual of the problem: "all" one has to do is find an algorithm for triangulating with a net that's all degree-4. I think if the surface is genus 0 this implies an octahedral "skeleton". Anyway, why wouldn't this work: Make a cubic cage with square mesh sides, put a light inside, insert in the target shape, then take the quads defined by the shadows on the surface? If not close enough, or produces non-planar quads, use (locally) finer mesh. (If the shape is too lumpy or floppy or the like first smoothly deform it to be rounder, project as above, then deform back...)
Yes, Marc's projection mechanism would work as an approximation, but would produce nearly 100% non-planar quadrilaterals. To some extent, I'm trying to understand the mathematics of planar quadrilaterals, so Warren's Euler formula provided me with significant insight about how one might go about proving the non-existence of some types of quadrangularization. Marc's octahedral skeleton is the dual of the cubical skeleton. I can take such a cubical skeleton and "push out" all of the faces, and then push out all of the resulting faces, in a recursive manner to produce a pretty decent approximation to the sphere. The original cubic skeleton has 6 faces. Since "pushing out" one face results in 5 new faces, one level of recursion gives us 6*5=30 total faces. Recursively pushing out n times results in 6*5^n total faces. Note that my "pushing out" algorithm approximates the sphere _from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere. BTW, my recursive mechanism provides a (new???) set of coordinates for every point on the surface of the sphere. My recursive mechanism also provides for a (new???) way of "projecting" (not really a mathematical projection in the usual sense) the surface of a sphere onto a set of reasonably compact planar quadrilaterals. Given all of the globe-mapping algorithms that have been suggested over the previous centuries, I'd be surprised if this one hasn't already been suggested at some point. Note that the little planar bits are (almost certainly) _not_ precisely perpendicular to a radius vector from the center of the sphere. This is a problem for computer graphics, because these little faces won't reflect correctly. We might have to optimize the choices involved in "pushing out", so as to make the planar quadrilaterals as compact as possible. Here the definition of "compact" might be to minimize the enclosing radius of the quadrilateral planar pieces at a given depth k of the recursion. At 12:58 PM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Well, how do we define "better"? On the one hand we seem to be talking about approximating surfaces ever more closely, but on the other we seem to be groping to converge on discovering some unstated intuitive definition.
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
Then you just get the dual of the problem: "all" one has to do is find an algorithm for triangulating with a net that's all degree-4.
I think if the surface is genus 0 this implies an octahedral "skeleton".
Anyway, why wouldn't this work:
Make a cubic cage with square mesh sides, put a light inside, insert in the target shape, then take the quads defined by the shadows on the surface?
If not close enough, or produces non-planar quads, use (locally) finer mesh.
(If the shape is too lumpy or floppy or the like first smoothly deform it to be rounder, project as above, then deform back...)
Oops! Without some mechanism for breaking the original cube's edges, we will get deep _creases_ at each one of these edges. We will also get smaller creases at each of the new edges as well. This would produce an extremely crinkled surface, including some surface elements that are nearly parallel with a radius vector from the center of the sphere. So we still need a rule to break the edges. Back to the drawing board... At 02:33 PM 1/6/2013, you wrote:
Yes, Marc's projection mechanism would work as an approximation, but would produce nearly 100% non-planar quadrilaterals.
To some extent, I'm trying to understand the mathematics of planar quadrilaterals, so Warren's Euler formula provided me with significant insight about how one might go about proving the non-existence of some types of quadrangularization.
Marc's octahedral skeleton is the dual of the cubical skeleton. I can take such a cubical skeleton and "push out" all of the faces, and then push out all of the resulting faces, in a recursive manner to produce a pretty decent approximation to the sphere.
The original cubic skeleton has 6 faces.
Since "pushing out" one face results in 5 new faces, one level of recursion gives us 6*5=30 total faces.
Recursively pushing out n times results in 6*5^n total faces.
Note that my "pushing out" algorithm approximates the sphere _from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere.
BTW, my recursive mechanism provides a (new???) set of coordinates for every point on the surface of the sphere.
My recursive mechanism also provides for a (new???) way of "projecting" (not really a mathematical projection in the usual sense) the surface of a sphere onto a set of reasonably compact planar quadrilaterals. Given all of the globe-mapping algorithms that have been suggested over the previous centuries, I'd be surprised if this one hasn't already been suggested at some point.
Note that the little planar bits are (almost certainly) _not_ precisely perpendicular to a radius vector from the center of the sphere. This is a problem for computer graphics, because these little faces won't reflect correctly.
We might have to optimize the choices involved in "pushing out", so as to make the planar quadrilaterals as compact as possible. Here the definition of "compact" might be to minimize the enclosing radius of the quadrilateral planar pieces at a given depth k of the recursion.
At 12:58 PM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Well, how do we define "better"? On the one hand we seem to be talking about approximating surfaces ever more closely, but on the other we seem to be groping to converge on discovering some unstated intuitive definition.
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
Then you just get the dual of the problem: "all" one has to do is find an algorithm for triangulating with a net that's all degree-4.
I think if the surface is genus 0 this implies an octahedral "skeleton".
Anyway, why wouldn't this work:
Make a cubic cage with square mesh sides, put a light inside, insert in the target shape, then take the quads defined by the shadows on the surface?
If not close enough, or produces non-planar quads, use (locally) finer mesh.
(If the shape is too lumpy or floppy or the like first smoothly deform it to be rounder, project as above, then deform back...)
I fail to see how this works given that the outside edges have never moved i.e. the edges of the whole mesh no matter how deep you go are still defined by just 4 corner points.... On 6 Jan 2013, at 22:33, Henry Baker wrote:
Yes, Marc's projection mechanism would work as an approximation, but would produce nearly 100% non-planar quadrilaterals.
To some extent, I'm trying to understand the mathematics of planar quadrilaterals, so Warren's Euler formula provided me with significant insight about how one might go about proving the non-existence of some types of quadrangularization.
Marc's octahedral skeleton is the dual of the cubical skeleton. I can take such a cubical skeleton and "push out" all of the faces, and then push out all of the resulting faces, in a recursive manner to produce a pretty decent approximation to the sphere.
The original cubic skeleton has 6 faces.
Since "pushing out" one face results in 5 new faces, one level of recursion gives us 6*5=30 total faces.
Recursively pushing out n times results in 6*5^n total faces.
Note that my "pushing out" algorithm approximates the sphere _from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere.
BTW, my recursive mechanism provides a (new???) set of coordinates for every point on the surface of the sphere.
My recursive mechanism also provides for a (new???) way of "projecting" (not really a mathematical projection in the usual sense) the surface of a sphere onto a set of reasonably compact planar quadrilaterals. Given all of the globe-mapping algorithms that have been suggested over the previous centuries, I'd be surprised if this one hasn't already been suggested at some point.
Note that the little planar bits are (almost certainly) _not_ precisely perpendicular to a radius vector from the center of the sphere. This is a problem for computer graphics, because these little faces won't reflect correctly.
We might have to optimize the choices involved in "pushing out", so as to make the planar quadrilaterals as compact as possible. Here the definition of "compact" might be to minimize the enclosing radius of the quadrilateral planar pieces at a given depth k of the recursion.
At 12:58 PM 1/6/2013, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com> Good, but we have a convex surface modelled by a non-convex surface. Can we do better?
Well, how do we define "better"? On the one hand we seem to be talking about approximating surfaces ever more closely, but on the other we seem to be groping to converge on discovering some unstated intuitive definition.
Duality. I note that a triangulated surface in which every vertex has exactly 4 edges coming out (I believe) is dual to a quadrangulated surface where each vertex meets exactly 3 planar quadrilaterals.
Then you just get the dual of the problem: "all" one has to do is find an algorithm for triangulating with a net that's all degree-4.
I think if the surface is genus 0 this implies an octahedral "skeleton".
Anyway, why wouldn't this work:
Make a cubic cage with square mesh sides, put a light inside, insert in the target shape, then take the quads defined by the shadows on the surface?
If not close enough, or produces non-planar quads, use (locally) finer mesh.
(If the shape is too lumpy or floppy or the like first smoothly deform it to be rounder, project as above, then deform back...)
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On 1/6/2013 2:33 PM, Henry Baker wrote:
Note that my "pushing out" algorithm approximates the sphere_from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere.
Is that true? What about the trapezoidal faces that are pushed out, not just the central square one? I don't think their vertices can all lie on a sphere. Brent Meeker
The vertices aren't the problem: they can all lie on the sphere. The problem is the faces that share one of the early edges become almost vertical (parallel with a radius vector). The surface becomes exceedingly crinkled. As I concluded in another post: we need to find a way to break the early edges, while preserving the properties that the quadrilaterals are planar and the faces all have 4 edges. While the corners are made up exclusively of only 3 faces coming together, this is not a requirement; perhaps in breaking an edge, we can construct corners with 4 (or more) faces. At 04:55 PM 1/6/2013, meekerdb wrote:
On 1/6/2013 2:33 PM, Henry Baker wrote:
Note that my "pushing out" algorithm approximates the sphere_from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere.
Is that true? What about the trapezoidal faces that are pushed out, not just the central square one? I don't think their vertices can all lie on a sphere.
Brent Meeker
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface. In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere. The construction I presented approaches the same volume, but not the same surface area. In fact, due to all the "crinkles", I think that the surface area is significantly larger than that of the sphere. Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere. At 05:36 PM 1/6/2013, Henry Baker wrote:
The vertices aren't the problem: they can all lie on the sphere. The problem is the faces that share one of the early edges become almost vertical (parallel with a radius vector). The surface becomes exceedingly crinkled.
As I concluded in another post: we need to find a way to break the early edges, while preserving the properties that the quadrilaterals are planar and the faces all have 4 edges. While the corners are made up exclusively of only 3 faces coming together, this is not a requirement; perhaps in breaking an edge, we can construct corners with 4 (or more) faces.
At 04:55 PM 1/6/2013, meekerdb wrote:
On 1/6/2013 2:33 PM, Henry Baker wrote:
Note that my "pushing out" algorithm approximates the sphere_from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere.
Is that true? What about the trapezoidal faces that are pushed out, not just the central square one? I don't think their vertices can all lie on a sphere.
Brent Meeker
No, that can't be right. What is the surface area of Marc's construction, which stuffs a sphere with cubes in a regular lattice, as the cubes get smaller & smaller ? At 06:05 PM 1/7/2013, Henry Baker wrote:
Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere.
The area of Marc's construction approaches 6*pi*r^2, while the surface of the sphere has area 4*pi*r^2. The surface is rough, as it has 50% more surface area than a smooth surface. Sketch of proof: 3 circular disks of radius r have area 3*2*pi*r^2 (counting both sides). If you arrange these 3 disks to intersect in 3 mutually perpendicular planes, this arrangement still has the same area. Now if you take Marc's spherical construction out of little cubes, you can remove the cubes, one by one, and each of these cube removal steps _does not change the total surface area_, until you reach the stage where you have 3 mutually perpendicular circular disks exactly one cube thick. (This area-preservation property comes from the fact that each cube removal removes 3 exposed sides of the removed cube, but exposes 3 new sides which were previously covered up.) Finally, if we allow the dimension of the cube to shrink towards zero, each disk's area (counting both sides) approaches 2*pi*r^2. QED. At 06:12 PM 1/7/2013, Henry Baker wrote:
No, that can't be right. What is the surface area of Marc's construction, which stuffs a sphere with cubes in a regular lattice, as the cubes get smaller & smaller ?
At 06:05 PM 1/7/2013, Henry Baker wrote:
Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere.
On 1/7/2013 6:05 PM, Henry Baker wrote:
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface.
In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere.
Wouldn't you want to require that for any distance d, the construction produce an approximating solid with quadrangular faces such that every point of the surface is within d of a point on a quadrangle and conversely every point on a quadrangle is within d of the surface? Brent Meeker
The construction I presented approaches the same volume, but not the same surface area. In fact, due to all the "crinkles", I think that the surface area is significantly larger than that of the sphere.
Marc LeBrun's construction also approaches the same _volume_ as the sphere, but his surface area is the same as the enclosing cube (I think), which is also significantly larger than the surface area of the sphere.
At 05:36 PM 1/6/2013, Henry Baker wrote:
The vertices aren't the problem: they can all lie on the sphere. The problem is the faces that share one of the early edges become almost vertical (parallel with a radius vector). The surface becomes exceedingly crinkled.
As I concluded in another post: we need to find a way to break the early edges, while preserving the properties that the quadrilaterals are planar and the faces all have 4 edges. While the corners are made up exclusively of only 3 faces coming together, this is not a requirement; perhaps in breaking an edge, we can construct corners with 4 (or more) faces.
At 04:55 PM 1/6/2013, meekerdb wrote:
On 1/6/2013 2:33 PM, Henry Baker wrote:
Note that my "pushing out" algorithm approximates the sphere_from the inside_, in the sense that the 8 vertices of the cube lie on the surface of the sphere, and each new "pushed out" face will also have all of its vertices on the surface of the sphere. Is that true? What about the trapezoidal faces that are pushed out, not just the central square one? I don't think their vertices can all lie on a sphere.
Brent Meeker
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For purposes of 3d graphics, a much more obvious property to approximate is the surface normal. The surface normal is used directly in graphic rendering to calculate reflectance for all but 100% "matte" surfaces. This immediately eliminates "crinkly" approximations. On 01/07/2013 09:05 PM, Henry Baker wrote:
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface.
In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere.
Yes, I'd love to do more accurate normals, but the first goal in this thread was to come up with a "quadrangulation" of the surface using planar quadrilaterals _only_ (no triangles). I still don't know how to quadrangulate a sphere such that both the volume & the surface area converge to reasonable values. I'm willing to have crinkles (including non-convexities), so long as they smooth out in the limit, which would also help solve the "normal" problem. At 05:11 PM 1/8/2013, John Aspinall wrote:
For purposes of 3d graphics, a much more obvious property to approximate is the surface normal. The surface normal is used directly in graphic rendering to calculate reflectance for all but 100% "matte" surfaces. This immediately eliminates "crinkly" approximations.
On 01/07/2013 09:05 PM, Henry Baker wrote:
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface.
In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere.
And are you asking that this be better in some sense than a quadrangulation that is produced by taking a triangulation and dividing each triangle into three quadrilaterals? Brent Meeker On 1/8/2013 5:41 PM, Henry Baker wrote:
Yes, I'd love to do more accurate normals, but the first goal in this thread was to come up with a "quadrangulation" of the surface using planar quadrilaterals _only_ (no triangles).
I still don't know how to quadrangulate a sphere such that both the volume& the surface area converge to reasonable values.
I'm willing to have crinkles (including non-convexities), so long as they smooth out in the limit, which would also help solve the "normal" problem.
At 05:11 PM 1/8/2013, John Aspinall wrote:
For purposes of 3d graphics, a much more obvious property to approximate is the surface normal. The surface normal is used directly in graphic rendering to calculate reflectance for all but 100% "matte" surfaces. This immediately eliminates "crinkly" approximations.
On 01/07/2013 09:05 PM, Henry Baker wrote:
I was thinking today that I need to make more precise my definition of "approximation" to a 3D surface.
In the case of a sphere, I think that 1) the limit of the series should have the same _volume_ as the sphere; and 2) the limit of the series should have the same _surface area_ as the sphere.
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Erm - I think a better solution would be split the initial quadrilateral into thirds - this also allows correction of the initial scale edges, --------- | | | | --------- | | | | --------- | | | | --------- On 6 Jan 2013, at 18:31, Marc LeBrun wrote:
="Henry Baker" <hbaker1@pipeline.com>
Can a smooth surface -- e.g., a hemisphere -- be approximated by a convex quadrangulated surface?
The answer is trivially yes -- the approximation is a cube -- but is there a way to quadrangulate it with smaller & smaller quadrilaterals to get a better & better approximation?
The answer is, again, trivially yes. Just keep stacking smaller aligned cubes into any gaps. Stop whenever your approximation is close enough. You will have a nice "cubic crystallized" model not only at the surface, but of the volume (that you can readily feed into your 3D printer if you want<;-).
Also keep in mind that, whereas any 3 points on a surface always define a planar triangle, any 4 points on a surface need not be co-planar; consider a saddle for example. So incremental triangularization schemes are inherently easier and more robust than successive quadrangularizations will be.
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron?
Although the intent seems intuitive, if all sorts of cases must be patched with a stream of new rules then I'd guess that it's probably because the problem to be solved isn't sufficiently precisely articulated yet.
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