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Main contributions The coefficients of the parametric form of P Q lie in an extension of which is, in general, of minimal degree. In this extension of, the parametric form does not contain nested radicals. The exact explicit form of the parametric solution has manageable size. Theorem: If P and Q intersect in more than 2 points, then there exists a rational number such that det(R( )) 0. Main contributions The coefficients of the parametric form of P Q lie in an extension of which is, in general, of minimal degree. In this extension of, the parametric form does not contain nested radicals. The exact explicit form of the parametric solution has manageable size. Theorem: If P and Q intersect in more than 2 points, then there exists a rational number such that det(R( )) 0. Robust Intersection of Two Quadric Surfaces Laurent Dupont - Daniel Lazard - Sylvain Lazard - Sylvain Petitjean Loria (Univ. Nancy 2, Inria, CNRS) - LIP6 (Univ. Paris 6) Applications: - Boundary evaluation (CSG to Brep) - Convex hull of quadric patches Applications: - Boundary evaluation (CSG to Brep) - Convex hull of quadric patches In affine space, find a simple ruled quadric R in the pencil R( ) = P- Q, real, such that is a solution of the degree 3 equation det(R u ( )) = 0. Compute the orthogonal transformation that sends R into canonical form: x²+ y²+ z+ =0 ( or =0) Parameterize R in its canonical frame. Ex: hyperbolic paraboloid x² - y² + z=0, , >0 (u,v) 2 Previous work: J. Levin (1976) We get a degree 2 equation in v: a(u)v²+b(u)v+c(u)=0 where a,b,c are degree 2 polynomials in u. Solve for v v (u) where u is such that b²-4ac 0. Substitute the solution of the degree 2 equation into the parametrization. Transform this solution from the local frame to the initial frame. In projective space, find a ruled quadric R in the pencil R( ) = P- Q, real, such that det(R( )) 0 and is rational. Using Gauss’ reduction method for quadratic forms, compute the linear transformation that sends R into canonical form: x²+ y²+ z²+ w²=0 Parameterize R in the local frame. Ex: signature (2,2) (corresponds to a hyperboloid of one sheet or a hyperbolic paraboloid, in affine space) x² + y² - z² - w²=0, , , , >0 (u,v),(t,s) 1 Compute the equation of Q in the local frame of R and substitute the parametrization into this equation. Our algorithm Advantages The use of a projective formalism simplifies the algorithm and its implementation (fewer types of quadrics, no infinite branches). The coefficients of the linear transformation and , , , are rational. These new parameterizations of canonical projective quadric surfaces are, in general, optimal in the number of radicals. We get a degree 2 homogeneous equation in s,t: a(u,v)s²+b(u,v)st+c(u,v)t²=0 where a,b,c are degree 2 homogeneous polynomials in u,v. Solve for (s,t) (s (u,v), t (u,v)) where (u,v) is such that b²-4ac 0. Given: P and Q, quadric surfaces (ellipsoids, hyperboloids, paraboloids,...) with rational coefficients. Problem: Find a parametric form for the intersection of P and Q. Given: P and Q, quadric surfaces (ellipsoids, hyperboloids, paraboloids,...) with rational coefficients. Problem: Find a parametric form for the intersection of P and Q. A rational can usually be found after computing an approximation of the roots of det(R( )). R has rational coefficients. Q has rational coefficients in the local frame.,, 2, 2 uvvuvu z y x T , vs,, vtusutvtusvsut w z y x T QRQP QRQP UNIVERSITE PIERRE & MARIE CURIE

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Fundamental Theorem of Algebra Every polynomial function of positive degree with complex coefficients has at least one complex zero.

Fundamental Theorem of Algebra Every polynomial function of positive degree with complex coefficients has at least one complex zero.

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