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Patrick Elliott <sel### [at] rraz com> wrote:
> ____
> \
> \
> \______
Please don't tell me you are using a variable-width font to write and read
news.
(It just wouldn't make any sense. By doing so you are assuming that
everyone else in the whole world is using the exact same font and font size
that you are.)
--
#macro M(A,N,D,L)plane{-z,-9pigment{mandel L*9translate N color_map{[0rgb x]
[1rgb 9]}scale<D,D*3D>*1e3}rotate y*A*8}#end M(-3<1.206434.28623>70,7)M(
-1<.7438.1795>1,20)M(1<.77595.13699>30,20)M(3<.75923.07145>80,99)// - Warp -
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Rune wrote:
>
> Le Forgeron wrote:
> > Ellipsoid is only a linear transformation of a
> > sphere, so it won't help when the normals do not meet.
>
> Does that logic apply?
>
> Three normals of a sphere will always meet in the same point. But three
> normals of a ellipsoid will not always meet in the same point. I'm not
> saying that an ellipsoid is a solution, but just questioning this
> particular argument.
>
Maybe you're right, and for some cases you can find an ellipsoid
which could accomodate the three normals (even if I have no idea of
how to find the ellipsoid parameters; using a unit sphere is easier
for the math).
But a linear transform cannot transform a positive curvature into
a negative one. So the ellipsoid won't solve all the cases.
Given also the troubling case where two normals are parallel (which
cannot be solve on someting like a sphere or even an ellipsoid),
I wonder if the curved triangle shouldn't be on a torus instead (*).
At least for the two parallel normals case, it seems that two vertices
would be on the very top circle of the torus with the third vertex somewhere
so that its normals is ok. Currently, I can only figure the math to
position the first vertex, the circle where the second vertex is and
where the third vertex might be.
Maybe using also the true normal of the initial triangle might help to
limit the position of the second and third vertices.
Once the triangle on the torus is defined, we still go back to the classical
transformation of three points from one space to another, nothing really difficult!
But maybe the torus is not even the right solution.
(*): The more I look at the problem, the more it make me think of an analogy with
conic curves (the 6 classical conic curves in 2D can all be view as the intersection
of a plane and an infinite double cone: ellipse, parabol, hyperbol, point, single and
double generating lines), which would means that to solve the really curved triangle,
you have first to find out which 3D objects should be used according to the
parameters...
Looks like an intersection of hyperplane with an hypercone, where the position of the
hyperplane
is made according to the parameter, isn't it ?
Only problem is that I can imagine in 3D, but 4D is a bit too much (excepted when it's
3D+Time,
which is not the case here)
--
Non Sine Numine
http://grimbert.cjb.net/
Etiquette is for those with no breeding;
fashion for those with no taste.
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> Maybe you're right, and for some cases you can find an ellipsoid
> which could accomodate the three normals (even if I have no idea of
> how to find the ellipsoid parameters; using a unit sphere is easier
> for the math).
> But a linear transform cannot transform a positive curvature into
> a negative one. So the ellipsoid won't solve all the cases.
>
> Given also the troubling case where two normals are parallel (which
> cannot be solve on someting like a sphere or even an ellipsoid),
> I wonder if the curved triangle shouldn't be on a torus instead (*).
>
> At least for the two parallel normals case, it seems that two vertices
> would be on the very top circle of the torus with the third vertex
somewhere
> so that its normals is ok. Currently, I can only figure the math to
> position the first vertex, the circle where the second vertex is and
> where the third vertex might be.
> Maybe using also the true normal of the initial triangle might help to
> limit the position of the second and third vertices.
>
> Once the triangle on the torus is defined, we still go back to the
classical
> transformation of three points from one space to another, nothing really
difficult!
> But maybe the torus is not even the right solution.
>
>
I would assume that for a curved triangle that had two parallel normals,
that the entire edge between those two normals would have the same normal.
for a torus, it is true that we can make a triangle that has two parallel
normals (ie, on the top of the torus) but it is quite easy to see that
between these two points the edge will curve (or even disappear in some
cases).
I don't think that simple primitives have the solution. There must be some
way of creating quadratic type surfaces without having a discrete mesh.
Perhaps isosurfaces could be accomodated, but I would think that would come
at great CPU expense.
-tgq
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> I don't think that simple primitives have the solution. There must be
some
> way of creating quadratic type surfaces without having a discrete mesh.
> Perhaps isosurfaces could be accomodated, but I would think that would
come
> at great CPU expense.
>
Just a furrther thought here.
I don't know how the vertices and normals are interpolated for a triangle,
but if it is a simple surface that can be described using some type of
equation, then perhaps that equation can be put into an isosurface function.
Anyone up for that?
-tgq
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On 15 Jul 2002 14:43:40 -0400, Warp <war### [at] tagpovrayorg> wrote:
> Please don't tell me you are using a variable-width font to write and read
> news.
Not to read no, but apparently to send. I use Opera and it has some quirks
I would like to beat out of it, but since I do my web comic reading, news groups
and the rest all in one shot from in it I have been reluctant to install another
program just to use it. Also my copy of MS LookOut has bugged and crashes
on POP3 accounts. Why it still works with the html hotmail I can't comprehend,
but...
In any case I didn't realize it was doing that. I'll see what if anything I can do
to fix it. :p
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In article <3d340ef3$1@news.povray.org>,
"TinCanMan" <Tin### [at] hotmailcom> wrote:
> I don't think that simple primitives have the solution. There must be some
> way of creating quadratic type surfaces without having a discrete mesh.
> Perhaps isosurfaces could be accomodated, but I would think that would come
> at great CPU expense.
Why try so hard to avoid using a triangle mesh? Any method you find will
probably be much more CPU hungry and have artifacts that are harder to
get rid of than faceting. The reason originally given for "curved smooth
triangles" was the dark artifact seen when the normal points away from
the camera, but there are other possible ways to fix that with flat
smooth triangles.
It seems like the best solution for curved triangles would be a cubic
surface tesselated into triangles, something like a triangular bezier
patch. It could even be tesselated on the fly to reduce memory
requirements.
Maybe subdivision surfaces would be a better solution.
--
Christopher James Huff <chr### [at] maccom>
POV-Ray TAG e-mail: chr### [at] tagpovrayorg
TAG web site: http://tag.povray.org/
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> Why try so hard to avoid using a triangle mesh? Any method you find will
> probably be much more CPU hungry and have artifacts that are harder to
> get rid of than faceting. The reason originally given for "curved smooth
> triangles" was the dark artifact seen when the normal points away from
> the camera, but there are other possible ways to fix that with flat
> smooth triangles.
> It seems like the best solution for curved triangles would be a cubic
> surface tesselated into triangles, something like a triangular bezier
> patch. It could even be tesselated on the fly to reduce memory
> requirements.
>
> Maybe subdivision surfaces would be a better solution.
I think another reason was that if not enough triangles are used, in
profile, the shape has a jagged edge rather than a smooth one. Even though
the shading is smooth, the outline still follows the polygons.
-tgq
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In article <3d35a378$1@news.povray.org>,
"TinCanMan" <Tin### [at] hotmailcom> wrote:
> I think another reason was that if not enough triangles are used, in
> profile, the shape has a jagged edge rather than a smooth one. Even though
> the shading is smooth, the outline still follows the polygons.
Still not a reason to avoid triangles altogether...if you use a
tesselated surface for the "curved triangles", you could still get the
straight edges small enough you can't tell the edge isn't really curved.
--
Christopher James Huff <chr### [at] maccom>
POV-Ray TAG e-mail: chr### [at] tagpovrayorg
TAG web site: http://tag.povray.org/
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On Wed, 17 Jul 2002 12:13:21 -0500, Christopher James Huff <chr### [at] maccom> wrote:
> In article <3d35a378$1@news.povray.org>,
> "TinCanMan" <Tin### [at] hotmailcom> wrote:
>
> > I think another reason was that if not enough triangles are used, in
> > profile, the shape has a jagged edge rather than a smooth one. Even though
> > the shading is smooth, the outline still follows the polygons.
>
> Still not a reason to avoid triangles altogether...if you use a
> tesselated surface for the "curved triangles", you could still get the
> straight edges small enough you can't tell the edge isn't really curved.
>
Only with an extraordinary number of triangles, and if you are using someone elses
model...
But yeah, if POV could be made aware of when such artifacts where visible and
automatically tessalated those surfaces sufficiently to erase the jagged surfaces,
that would
be nice. The problem is that no program I know of does and not everyone can photoshop
them
away without making things worse. The dark artifacts 'may' be fixed with the right
changes,
but squarish 'curves' will only be solved by either replacing them with something that
can
be CSGed, further tessalation or some alternate solution. In general, the equation is:
Realism = Primatives Used / (Meshes / Complexity) and in that equation even one mesh
no
matter how complex 'will' effect the overall realism, at least until you have the
resources and
time of companies like Dreamworks and can make the complexity so high that the result
of the bottom half becomes 0.xxxx. I doubt most of us have those kinds of resources.
Neadless to say a better solution is definitely needed. ;)
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In article <1103_1026930387@news.povray.org>,
Patrick Elliott <sel### [at] rrazcom> wrote:
> Only with an extraordinary number of triangles, and if you are using someone
> elses model...
The tesselation could be done at render time without too much drop in
speed, and the number of triangles wouldn't be so "extraordinary". And
it could work with any mesh, I don't know where you got the "someone
elses model" limitation.
> But yeah, if POV could be made aware of when such artifacts where
> visible and automatically tessalated those surfaces sufficiently to
> erase the jagged surfaces, that would be nice. The problem is that no
> program I know of does and not everyone can photoshop them away
> without making things worse.
I don't think it is such a problem...just increase the amount of
tessellation for triangles with a high curvature and "convex" triangles
where the normal for a flat triangle with the same points would be at a
high angle to the incoming ray. Maybe also adjust the tesselation
fineness across the triangle. There might be trouble with "tearing" or
"cracks", but there might be a way to prevent that, and it might not be
a big problem.
> The dark artifacts 'may' be fixed with the right changes, but
> squarish 'curves' will only be solved by either replacing them with
> something that can be CSGed, further tessalation or some alternate
> solution.
If you are using a mesh anyway, further tesselation seems like a perfect
solution.
--
Christopher James Huff <chr### [at] maccom>
POV-Ray TAG e-mail: chr### [at] tagpovrayorg
TAG web site: http://tag.povray.org/
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