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Apollonius Meets Pythagoras1 Apollonius meets Pythagoras1 Ching-Shoei Chiang* and Christoph M. Hoffmann† * Soochow University, Taipei, Taipei, R.O.C. E-mail: [email protected] Tel/Fax: +886-2-2311-1531 ext. 3801 † Purdue University, West Lafayette, IN 47906, USA E-mail: [email protected] Tel: 1-765-494-6185 Abstract—Cyclographic maps are an important tool in rational Bézier curve. Section 4 illustrates the Apollonius Laguerre geometry that can be used to solve the problem of problem of 3 primitives, including ray, cycle, and oriented PH Apollonius. The cyclographic map of a planar boundary curve is curve. We conclude with results in the final section. a ruled surface that we call -map. A PH curve is a free form curve whose tangent and normal are polynomial. This implies that its cyclographic map has a parametric form. This paper considers the medial axis transform (MAT) of a PH curve with a II. DEFINITIONS AND THEOREMS ray and the MAT of a PH curve with a cycle. We transform the Pythagorean Hodograph(PH) has a property that is closely MAT finding problem into a surface intersection problem. Here, linked to the its cyclographic map. If a 2D curve has the PH the intersection curve is representable in rational Bezier form. property, that is, if its tangent is a polynomial, then the Furthermore, the generalized Apollonius problem concerning rays, cycles, and oriented PH curves is introduced. cyclographic map for the curve has a polynomial form. We define some key terms now, including the Pythagorean Hodograph curve and the cyclographic map. I. INTRODUCTION A. Planar PH boundary Cyclographic maps are widely used in CAGD and other areas The hodograph of a parametric curve is the locus described [3,12]. They were developed in classical Descriptive by the first derivatives of the curve. If the hodograph Geometry[10] and used to solve the problem of Apollonius, x'(t) x(t) and, more generally, as a device for studying distance maps, r'(t) of a curve r(t) satisfies y'(t) y(t) surface curvature, and other basic geometric properties. 2 2 2 Cyclographic maps for a planar PH curve have a parametric (t) x (t) y (t) where (t) is a polynomial, then r(t) is representation. However, the shape of a PH curve is not so a Pythagorean Hodograph curve (PH curve) [5]. The PH easy to imagine given its parameter values. In past work, curve has many properties such as: polynomial arc length, cyclographic maps have been approximated using Hermite rational offset and rational curvature. interpolation to the resulting PH curve [22,23,24]. Here, we consider instead one curve -- with a relatively high degree. To Definition 1:[7] (Complex representation) construct the degree 2n-1 PH curve, we can choose n control A PH curve in the complex plane is a complex valued points freely, but must derive the remaining n-1 control polynomial curve of the form r(t) x(t) iy(t) There must points[1], the up to n+1 additional control points can also be exist a complex polynomial (t) u(t) iv(t) , with obtained by some natural restrictions on data set[16]. After gcdu(t),v(t) 1 , such that the hodograph is finding the parametric form of a PH curve, we can derive the 2 2 2 Cyclographic map for it, considering the curve to bound a r(t) (t) u (t) v (t) i2u(t)v(t) . planar region. In this article, we employ the cyclographic maps for PH boundaries and transform finding the MAT of Although the hodograph is defined in the complex plane, we the region into a surface/surface intersection problem. We can map the real and imaginary parts into the Cartesian x and also solve how to find the MAT of a PH curve with a ray and y coordinates, respectively. the MAT of a PH curve with a cycle. x(t) Thus, for a PH curve r(t) , the hodograph can be y(t) The material is presented in 5 sections. Section 1 describes u 2 (t) v2 (t) the purpose of this paper. Section 2 states some definitions represented as r'(t) , such that the polynomial and theorems that have been derived by others, and that are 2u(t)v(t) useful for this paper. Section 3 derives the theorems needed 2 for solving the MAT for a PH curve with a ray or cycle, (t) u2 (t) v2 (t) 2u(t)v(t) 2 u2 (t) v2 (t) . showing that the intersection curve representing the MAT is a 1 Work supported in part by NSC Grant NSC 97-2212-E-031-002, NSF CPATH CCF-0722210, and DOE DE-FG52-06NA26290. The following is straightforward. Definition 2: (The -maps for a Ray) [13] An oriented line is called a ray. For a ray L: Ax+By+D=0 in the xy-plane, the vector (A,B) is the normal vector on the left Theorem 1: (σ(t), x (t) and y (t) in Bézier form) [1] hand side of the ray, so the orientation of the ray is thereby Given two polynomial u(t) and v(t) of degree μ in Bernstein- assigned. The -map of the ray L is the plane (L): Ax By A2 B2 z D 0 and is a plane in 3-space. Bézier form as u(t) B (t)u and v(t) B (t)v , we can i i i i i0 i0 construct the hodograph for a PH curve Definition 3: (The -maps for a Cycle) [13] An oriented circle is called a cycle. For a cycle 2 2 x(t) u (t) v (t) 2 2 2 2 2 , (t) u (t) v (t) . Then we can find c: (x a) ( y b) r in the xy-plane, r is called the y(t) 2u(t)v(t) oriented radius. If r>0, the cycle is counterclockwise, and 2 2 2 clockwise for r < 0. The -map of the cycle c is the cone (c): (t) B2 (t)H , x(t) B2 (t)I , y(t) B2 (t)J . k k k k k k (x a)2 (y b)2 (z r)2 0 in 3-space. k0 k0 k0 Let the control points for σ(t), x(t) and y(t) be Hk, Ik and Jk respectively: For a ray L in the xy-plane, (L) is a plane which inclined 45 k degrees to the xy-plane, through L. For a cycle c in the xy- , plane, (c) is a right circular cone whose generators are Hk i,ki (uiuki vivki ) , k=0…2μ i0 inclined 45 degrees to the xy-plane. (see figure 1a) k I , (u u v v ) , k=0…2μ k i,k i i k i i k i i0 k J , (u v v u ) , k=0…2μ k i,k i i k i i k i i0 (a) (c) (b) (B(t)) C mC n Figure 1 : The -maps for the 2D boundary where m,n i j , if i<0 or j<0, n,m 0 i, j mn i, j Ci j Definition 4: (Rotated normal vector fields of PH boundary) [3] Theorem 2: (PH curve in Bézier form) [1] X (t) 2 2 I k For a 2D PH boundary B(t) Y (t) in the xy-plane, the For the hodograph r(t) B (t) in Theorem 1, the k J 0 k0 k corresponding PH curve is a polynomial curve of degree orientation is induced by the parameterization. The normal 2 1 . Given the point p0, we can construct the PH curve in Y (t) 2 1 vector field on the left hand side is X (t) . Because of Bernstein Bézier form: B(t) B2 1(t) p , the control i i 0 i0 Y (t) points pk are: 2 2 (t) X (t) Y (t) , the vector field V (t) X (t) is 1 Ik pk1 pk , for k = 0... 2 (t) 2 1 J k inclined 45 degrees to the xy-plane. We call V(t) the rotated normal vector field of B(t). B. Cyclographic maps Cyclographic maps have been used since 1929 till now Theorem 3: (Rotated normal vector fields in Bézier form) [10,11,12,13]. The Apollonius problem for three circles is to According to Theorem 1, given two polynomial u(t) and v(t) find the 8 tangent circles to these three circles. When the three of degree μ, we can derive the rotated normal vector field of a 2 circles are oriented, we can divide the Apollonius problem PH curve in Bernstein Bézier form as: V (t) B2 (t)q . into 4 sub-problems, and each determining two tangent k k circles[25,26]. Using a similar approach, we want to solve the k0 Apollonius problem for three geometric entities chosen from The control points qk are: lines, circles, and PH curves. In this paper, we treat lines, J k circles, and PH curves as rays (oriented lines), cycles qk I k , for k = 0... 2 (oriented circles) and oriented PH curves. We also call the H k cyclographic map of an entity the r-map of the entity. The B(t) and V(t) of Definition 4 define a ruled surface C. The intersection of a ruled surface and a plane S(s,t)=B(t)+sV(t) which has inclination angle of 45 degrees The problem (a) “The one sided offset curve of a PH curve” with xy-plane. The ruled surface is the r-map for B(t) with and the problem (b) “The MAT of a PH curve and a ray” are formula(see figure 1b)[3]: about the intersection of the PH curve’s -map intersect with a plane.
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