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On Sat, 9 Dec 2000 18:53:44 +0100, Rune wrote:
>How do I make patches fit seamlessly together when the number of patches
>that has corners in the same point is not 4?
>
>The usual trick of having the points aligned in lines doesn't seem to apply,
>since it only makes sense when there's 4 patches sharing a corner.
You just need to make sure that the control points are all coplanar. That is,
if you have, say, three patches joined together in a corner like so:
x
\ / \ /
\ / \ /
* A *
|\ /|
\| \ / |/
xC O Bx
|\ | /|
\|/
*
|
Assume that the corner is O, and the intersections marked by *s are on the
edges of two patches. (The lines going through points marked X lie inside
the joined patches, which are not completely represented here. If the
picture isn't entirely clear, ask me and I'll try to throw together a POV
scene that describes what I'm talking about.)
What you have to do to make sure there aren't any "dimples" at O is make sure
that the three points marked with *s are coplanar with the point marked with
an O. To make sure you don't get creases, you have to make sure the points
marked with an x follow the usual rule for seamless patch joinery: if an x,
a *, and an x are adjacent, they should be in a straight line.
So, for your sphere made of six patches, you'd put each set of *'d points
somewhere on a face of the octahedron whose faces are centered on the vertices
of the cube.
The solution does not generalize, however. You can have four patches come
together seamlessly at a point without the four adjacent edge points being
coplanar, but there are other more difficult constraints on that solution.
You can be sure, though, that if the adjacent points are all coplanar, you
won't get a dimple. That is, having coplanar points is sufficient for a
smooth solution, but not necessary except in the case of three patches.
--
Ron Parker http://www2.fwi.com/~parkerr/traces.html
My opinions. Mine. Not anyone else's.
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