Talk:Bézier surface

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There's a lot more that can be said about this topic; I know enough to know I have forgotten most of it. Basis vectors, vector spaces, continuity, tensors, topology... -- Karada 12:37, 31 Jul 2003 (UTC)


From memory:

Note that this property is not in general true of arbitary lines in (u,v) space.

I think that's true: can someone confirm it? -- Karada 13:06, 31 Jul 2003 (UTC)

Concerning your last two paragraphs: See "Bezier and B-Spline Techniques" (Prautzsch et al) pp 166 for conversion between tensor product and triangular patches. Your are right that a m times n patch can be represented as a single m+n triangular patch, since both represent polynomial surfaces of the same degree. I think you just have to transform the input domain from square to barycentric coordinates, though I am not sure how you would do that. An idea how you might convert the control points for 1x1 TP -> 2x2 triangular is represented here: http://www.mmweg.rwth-aachen.de/~arne.schmitz/download/tensor.png -- dunno if that is correct, but I think it makes sense. Root 42 20:26, 29 Jan 2004 (UTC)

[edit] Definition of Bézier surface

I modified the definition of a Bézier surface. Strictly speaking, the poles are defined in an affine space and are not vectors. Typically, a vector does not depend on the choosing of the origin, but a point does depend. Also, I specified the range of the parameters (u,v), since it can be convenient to change this range when converting into orthogonal polynomials basis like Legendre or Tchebychev. I did not mention this in my editing.


[edit] ================

I'm probably not doing this right, but I'm just learning about wikipedia. In the sentence

All u = constant and v = constant lines in the (u, v) space, and, in particular, all four edges of the deformed (u, v) unit square are Bézier curves.

you really mean to say that the _images_ of these lines are Bezier curves, don't you? This is what people mean by a Bezier curve; the image. (Notwithstanding that in differential geometry 'curve' technically means the mapping itself. And yes, these lines (u=const, v=const)in the domain itself are also trivially Bezier curves, but no way do you mean that -- you write too well for any such unbridled pedantry)

Finn 68.55.44.143 01:31, 9 January 2007 (UTC)

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