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The Extended Oloid and Its Inscribed Quadrics

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The Extended Oloid and Its Inscribed Quadrics The extended oloid and its inscribed quadrics Uwe B¨asel and Hans Dirnb¨ock Abstract The oloid is the convex hull of two circles with equal radius in perpen- dicular planes so that the center of each circle lies on the other circle. It is part of a developable surface which we call extended oloid. We determine the tangential system of all inscribed quadrics Qλ of the extended oloid O where λ is the system parameter. From this result we conclude parameter equations of the touching curve Cλ between O and Qλ, the edge of regression R of O, and the asymptotes of R. Properties of the curves Cλ are investigated, including the case that λ ! ±∞. The self-polar tetrahedron of the tangential system Qλ is obtained. The common generating lines of O and any ruled surface Qλ are determined. Furthermore, we derive the curves which are the images of Cλ and R when O is developed onto the plane. Mathematics Subject Classification: 51N05, 53A05 Keywords: oloid, extended oloid, developable, tangential system of quadrics, touching curve, edge of regression, self-polar tetrahedron, ruled surface 1 Introduction The oloid was discovered by Paul Schatz in 1929. It is the convex hull of two circles with equal radius r in perpendicular planes so that the center of each circle lies on the other circle. The oloid has the remarkable properties that it develops its entire surface while rolling, and its surface area is equal to 4πr2. The surface of the oloid is part of a developable surface. [2], [8] In the following this developable surface is called extended oloid. Ac- cording to [2, pp. 105-106], with r = 1 the circles can be defined by arXiv:1503.07399v3 [math.MG] 14 Apr 2015 n 3 2 1 2 o kA := (x; y; z) 2 R x + y + 2 = 1 ^ z = 0 ; n 3 1 2 2 o kB := (x; y; z) 2 R y − 2 + z = 1 ^ x = 0 : In this case we denote the extended oloid by O (see Fig. 1). Now we introduce homogeneous coordinates x0; x1; x2; x3 with x x x x = 1 ; y = 2 ; z = 3 : x0 x0 x0 1 Fig. 1: The extended oloid O, the circles kA, kB (dashed lines), and the edge of regression R (solid lines) in the box −2:5 ≤ x; y; z ≤ 2:5 Then the real projective space is given by 4 P3(R) = [x0; x1; x2; x3] j (x0; x1; x2; x3) 2 R n f0g ; where [x0; x1; x2; x3] = [y0; y1; y2; y3] if there exist a µ 2 R n f0g such that xj = µyj for j = 1; 2; 3. Should it prove necessary, complex coordinates will be used instead of the real ones. For the description of the corresponding projective circles KA and KB to kA and kB, respectively, we write KA := [x0; x1; x2; x3] 2 P3(R) 'A(x0; x1; x2; x3) = 0 ^ x3 = 0 ; KB := [x0; x1; x2; x3] 2 P3(R) 'B(x0; x1; x2; x3) = 0 ^ x1 = 0 with 2 2 2 'A(x0; x1; x2; x3) = 3x0 − 4x0x2 − 4x1 − 4x2 ; 2 2 2 'B(x0; x1; x2; x3) = 3x0 + 4x0x2 − 4x2 − 4x3 : 2 Inscribed quadrics Lemma 1. The dual figures to KA and KB are 2 2 2 KbA = [u0; u1; u2; u3] 2 P3(R) 4u0 − 4u0u2 − 4u1 − 3u2 = 0 and 2 2 2 KbB = [u0; u1; u2; u3] 2 P3(R) 4u0 + 4u0u2 − 3u2 − 4u3 = 0 ; 2 respectively, where u0; u1; u2; u3 are homogeneous plane-coordinates. KbA and KbB are elliptic cylinders. Their respective non homogeneous equations are 2 2 2 u v + 3 2 + 2 = 1 ^ −∞ < w < 1 ; 2 4 p 3 3 and 2 2 2 v − 3 w −∞ < u < 1 ^ 2 + 2 = 1 : 4 2 3 p 3 Proof. Following [3, pp. 41-42], [4, pp. 160, 164-165], we determine KbA. With @'A µui = ; i 2 f0; 1; 2; 3g ; @xi we get µu0 = 6x0 − 4x2 ; µu1 = −8x1 ; µu2 = −4x0 − 8x2 ; µu3 = 0 : Solving this system of linear equations for x0, x1, x2 delivers 1 1 1 1 3 x = µ u − u ; x = µ − u ; x = µ − u − u : 0 8 0 16 2 1 8 1 2 16 0 32 2 So we get 1 0 = u x + u x + u x + u x = µ 4u2 − 4u u − 4u2 − 3u2 ; 0 0 1 1 2 2 3 3 32 0 0 2 1 2 hence 2 2 2 KbA := [u0; u1; u2; u3] 2 P3(R) 4u0 − 4u0u2 − 4u1 − 3u2 = 0 : Analogously, one finds KbB from KB. With u0 = 1, u1 = u, u2 = v, u3 = w, the non homogeneous polynomials for KbA and KbB follow immediately. 2 2 2 2 Remark 1. The equations 4u0 − 4u0u2 − 4u1 − 3u2 = 0 and 4u0 + 4u0u2 − 2 2 3u2 − 4u3 = 0 of KbA and KbB, respectively, were already given in [2, p. 115]. Theorem 1. The inscribed quadrics of the extended oloid O are given by 3 Qλ = (x; y; z) 2 R fλ(x; y; z) = 0 with 2 x2 y − λ + 1 z2 f (x; y; z) = + 2 + − 1 : λ 1 − λ 1 − λ + λ2 λ 3 Proof. For abbreviation we put 2 2 2 2 2 2 F0(¯u) := 4u0 − 4u0u2 − 4u1 − 3u2 ;F1(¯u) := 4u0 + 4u0u2 − 3u2 − 4u3 withu ¯ := [u0; u1; u2; u3]. Then Fbλ := fu¯ 2 P3(R) j Fλ(¯u) = 0g with Fλ(¯u) := (1 − λ) F0(¯u) + λ F1(¯u) defines a tangential system of quadrics in plane-coordinates (cp. [6, p. 253]). Now we shall determine the point-coordinate representation Fλ of Fbλ. Due to duality (see [4, p. 163]) we have @Fλ(¯u) @F0(¯u) @F1(¯u) σxj = = (1 − λ) + λ ; j = 0; 1; 2; 3: @uj @uj @uj The calculation of the partial derivatives yields σx0 = (1 − λ)(8u0 − 4u2) + λ(8u0 + 4u2) = 8u0 + 4(2λ − 1)u2 ; σx1 = −8(1 − λ)u1 ; σx2 = (1 − λ)(−4u0 − 6u2) + λ(4u0 − 6u2) = 4(2λ − 1)u0 − 6u2 ; σx3 = − 8λu3 : Solving this system of linear equations for u0; : : : ; u3 delivers 3x − 2(1 − 2λ)x x u = σ 0 2 ; u = −σ 1 ; 0 32(1 − λ + λ2) 1 8(1 − λ) (1 − 2λ)x + 2x x u = −σ 0 2 ; u = −σ 3 : 2 16(1 − λ + λ2) 3 8λ So we find 3 2 2 2 2 ! X σ 4x 4 x + (1 − 2λ)x0x2 − 3x 4x 0 = x u = − 1 + 2 0 + 3 i i 32 1 − λ 1 − λ + λ2 λ i=0 2 1 2 2 ! σ x x2 + − λ x0 x = − 1 + 2 + 3 − x2 ; 8 1 − λ 1 − λ + λ2 λ 0 and therefore ~ Fλ = [x0; x1; x2; x3] 2 P3(R) fλ(x0; x1; x2; x3) = 0 with 2 1 2 2 x x2 + − λ x0 x f~ (x ; x ; x ; x ) = 1 + 2 + 3 − x2 : (1) λ 0 1 2 3 1 − λ 1 − λ + λ2 λ 0 4 Finally, we write the representation of Fλ in non homogeneous coordinates x; y; z with x0 = 1, x1 = x, x2 = y, x3 = z as 3 Qλ = (x; y; z) 2 R fλ(x; y; z) = 0 ; where 2 x2 y − λ + 1 z2 f (x; y; z) = + 2 + − 1 : λ 1 − λ 1 − λ + λ2 λ Now we classify the quadrics Qλ with real parameter λ in Euclidean space. We start with the case that λ tends to ±∞. One finds lim fλ(x; y; z) = 0 : λ→±∞ So we consider λ · fλ instead of fλ and find x2 x2 z2 lim λ · = lim = −x2 ; lim λ · = z2 ; λ→±∞ λ→±∞ 1 λ→±∞ 1 − λ λ − 1 λ 2 ! y − λ + 1 λ y2 + y − 3 − 2λ2y lim λ · 2 − 1 = lim 4 λ→±∞ 1 − λ + λ2 λ→±∞ 1 − λ + λ2 1 y2 + y − 3 − 2y = lim λ 4 = −2y ; λ→±∞ 1 1 λ2 − λ + 1 hence 2 2 lim λ fλ(x; y; z) = −x + z − 2y : λ→±∞ In order to abbreviate notation we put a2 := j1 − λj, b2 := 1 − λ + λ2 > 0 for 2 every λ 2 R, c := jλj. So we have the quadrics Qλ in the following table: λ = −∞ x2 − z2 + 2y = 0 Hyperbolic paraboloid 2 x2 y − λ + 1 z2 3 λ < 0 + 2 − = 1 Hyperboloid of one sheet R a2 b2 c2 2 1 2 λ = 0 x + y + 2 = 1 ^ z = 0 Circle kA 2 x2 y − λ + 1 z2 0 < λ < 1 + 2 + = 1 Ellipsoid a2 b2 c2 1 2 2 λ = 1 y − 2 + z = 1 ^ x = 0 Circle kB 2 y − λ + 1 z2 x2 3 λ > 1 2 + − = 1 Hyperboloid of one sheet R b2 c2 a2 λ = 1 x2 − z2 + 2y = 0 Hyperbolic paraboloid 5 A point of the circle kA is given by A = α1(t); α2(t); α3(t) with 1 α (t) = sin t ; α (t) = − − cos t ; α (t) = 0 : 1 2 2 3 There are two points B1, B2 of the circle kB which have common generating lines AB1 and AB2, respectively, with A: B1 = β1(t); β2(t); β3(t) ;B2 = β1(t); β2(t); −β3(t) ; where p 1 cos t 1 + 2 cos t β (t) = 0 ; β (t) = − ; β (t) = 1 2 2 1 + cos t 3 1 + cos t (see [2, pp. 106-107]). Hence, for fixed t 2 [−2π=3; 2π=3], parametric func- tions of a line AB1 are !i(m; t) := (1 − m)αi(t) + mβi(t) ; m 2 R ; i = 1; 2; 3 : One finds ! (m; t) = (1 − m) sin t; 9 1 > > 2(m − 1) cos2 t + (2m − 3) cos t + 2m − 1 > ! (m; t) = ; = 2 2(1 + cos t) (2) p > m 1 + 2 cos t > ! (m; t) = : ;> 3 1 + cos t It follows that ) 1) x = !1(m; t) ; y = !2(m; t) ; z = !3(m; t) ; 2π 2π t 2 − ; ; m 2 R ; 2) x = !1(m; t) ; y = !2(m; t) ; z = −!3(m; t) ; 3 3 are the parametric equations of all generating lines of O, hence a parametri- sation of O.
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