By Stephen Mann

During this lecture, we examine Bézier and B-spline curves and surfaces, mathematical representations for free-form curves and surfaces which are universal in CAD platforms and are used to layout airplane and autos, in addition to in modeling programs utilized by the pc animation undefined. Bézier/B-splines characterize polynomials and piecewise polynomials in a geometrical demeanour utilizing units of keep watch over issues that outline the form of the outside. the first research device utilized in this lecture is blossoming, which supplies a chic labeling of the keep an eye on issues that permits us to research their homes geometrically. Blossoming is used to discover either Bézier and B-spline curves, and specifically to enquire continuity homes, switch of foundation algorithms, ahead differencing, B-spline knot multiplicity, and knot insertion algorithms. We additionally examine triangle diagrams (which are heavily concerning blossoming), direct manipulation of B-spline curves, NURBS curves, and triangular and tensor product surfaces.

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**Extra resources for A blossoming development of splines**

**Sample text**

All derivatives are 0 and the curve is of zero length over a nonzero length interval in the domain) and we can introduce a cusp in the curve. (a) Suppose that we have the knot sequence (0, 1, 2, 3, 4, 5, 6, 7) for a cubic B-spline f , and further suppose that f (2, 3, 4) = f (3, 4, 5). The theorem on continuity tells us that this curve should be C 2 at F(3). Is the curve geometrically C 2 at F(3)? If so, support your statement. If not, give a counterexample. (b) Now suppose that f (1, 2, 3) = f (2, 3, 4) = f (3, 4, 5).

Thus, the location of f (4, 5, 6, 7) can be calculated from the control points of the first segment. Once these control points are fixed, the location of f (4, 5, 6, 7) from these control points is determined. But in a B-spline, we are free to place the gray point f (4, 5, 6, 7) in the diagram anywhere we please. And, as seen in the diagram, we can place it at a location different from the one we would calculate the control points of the first segment. If we were to calculate the B´ezier control points of the two segments, we would see that they meet with C 2 continuity at t = 4.

0¯ , δ, . . , δ ) f ∗ (0, (n − i)! n−i i Now, we have the following: n! n! ¯ . . , u¯ , δ, . . , δ ) = f ∗ (u, (n − j )! (n − j )! n− j n! (n − j )! n− j n! (n − j )! n− j n− j j = = n− j = k=0 ( j) F k=0 k n− j −k j j +k (n − j )! (n − j − k)! (n − j − k)! n! (0) k! n− j −k n− j ¯ . . , 0¯ , δ, . . , δ )u k f ∗ (0, k k=0 k=0 ( j +k) = F (u) n− j ¯ . . , 0¯ , δ, . . , δ )u k f ∗ (δ, . . 4. ) = Note that this is multilinear: each term has either u i or wi as a linear term. Thus, α f ∗ (u, ¯ .