ON THE RIGIDITY OF POLYGONAL MESHES ... - TU Wien-Geometry

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a flat pose of this four-sided pyramid with congruent faces we can even .... An interesting continuously flexible quadrilateral mesh dates also back to Kokot-.
South Bohemia Mathematical Letters Volume 19, (2011), No. 1, 6-17.

ON THE RIGIDITY OF POLYGONAL MESHES HELLMUTH STACHEL Dedicated to Prof. Pavel Pech at the occasion of his 60th birthday Abstract. A polygonal mesh is a connected subset of a polyhedral surface. We address the problem whether the intrinsic metric of a mesh, i.e., its development, can determine the exterior metric. If this is the case then the mesh is rigid. Among the non-rigid cases even flexible versions are possible. We concentrate on quadrangular meshes and in particular on a mesh with a flat pose in which the quadrangles belong to a tesselation. It is proved that this mesh admits a self-motion and that all its flexions represent discrete models of cylinders of revolution. These flexions can be generated from a skew line-symmetric hexagon by applying iterated coaxial helical motions.

Introduction In the following we understand under a polygonal mesh a connected subset of any polyhedral surface in the Euclidean 3-space E3 . This means, a polygonal mesh is a surface consisting of planar polygonal faces, edges and vertices. The edges are either internal when they are shared by two faces, or they belong to the boundary of the mesh. The term combinatorial structure of the mesh stands for the list of faces and the identification of those pairs (fi , fj ) of faces which share an internal edge eij . In this sense a polyhedron is a polygonal mesh with internal edges only. Suppose that this mesh undergoes a transformation which acts on the faces as isometry and preserves their planarity. We call this mesh rigid when under this transformation also all dihedral angles between adjacent faces are preserved. The question whether the intrinsic metric of the mesh determines its spatial shape uniquely or not is also important for many engineering applications, e.g., for mechanical or constructional engineers, for biologists in protein modelling or for the analysis of isomers in chemistry. Polygonal meshes, in particular quadrangular meshes, play an important role in discrete differential geometry and in new architecture where they serve as a discrete model of freeform surfaces. In the following we present different kinds of rigidity and we characterize the flexions of Kokotsakis’ flexible quadrangular tesselation of the plane. It should be noted that we only focus on geometric aspects of flexibility. We do not treat technical aspects like stiffness of faces and edges or clearances along hinges.

Received by the editors September 15, 2011. 1991 Mathematics Subject Classification. 51M20, 52B10, 53A17. Key words and phrases. flexible polyhedra, snapping polyhedra, Kokotsakis meshes, origami mechanisms. 6

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1. The definitions of rigidity Definition 1. A polyhedron or a polygonal mesh is called “globally rigid” when its development (unfolding) defines its spatial shape uniquely — apart from movements in space. The development of the polygonal mesh defines its intrinsic metric, i.e., the true shape of the faces and the combinatorial structure. Global rigidity means that the intrinsic metric defines the exterior metric in space including the dihedral angle ϕij at each internal edge eij shared by fi and fj . Conversely, the development together with all dihedral angles defines the spatial shape completely as we can assemble the mesh face by face, provided angle ϕij is signed with respect to a prescribed orientation of edge eij . Of course, the dihedral angles cannot be given independently; they must be compatible with the intrinsic metric.

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Figure 1. Two realizations of the same net. From everybody’s experience with assembling cardboard model of cubes, prisms or pyramids during school-days one tends to conjecture that each polyhedron is globally rigid. In fact, it is true, e.g., for a three-sided pyramid (tetrahedron) or for a cube or — more generally — for all polyhedra where at each vertex exactly three faces are meeting. However, the example in Fig. 1 shows two incongruent realizations with the same development. The polyhedron on the left-hand side is convex; it is built from a cube where the top face is replaced by pyramid. The polyhedron on the right-hand side contains edges along which the interior dihedral angle is > π. When the height of the attached pyramid is sufficiently small, we can transform the convex version into the concave one by applying a slight force to the apex of the pyramid. In this case we speak of “snapping” polyhedra; we can vary the spatial shape between two possibilities when admitting small deformations in between, e.g., by slight bending of faces and edges. Theoretically, both realizations are locally rigid according to the following definition. Definition 2. A polygonal mesh is called “locally rigid”, if its intrinsic metric admits no other realization with dihedral angles sufficiently close to that of the given one. Surprisingly, there are examples of polyhedra where the development admits even infinitely many incongruent realizations. Definition 3. A polygonal mesh is called flexible if there is a continuous family of mutually incongruent meshes sharing the intrinsic metric. In this case the mesh admits a self-motion; each pose obtained during this self-motion is called a flexion.

HELLMUTH STACHEL

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Figure 2. The regular octahedron and its re-assembled continuously flexible versions.

In Fig. 2 a trivial example of a flexible polyhedron is displayed. The intrinsic metric originates from that of a regular octahedron. We can re-assemble this polyhedron by putting one four-sided pyramid into the other. This gives a twofold covered quadratic pyramid without basis, which of course admits a self-motion. From a flat pose of this four-sided pyramid with congruent faces we can even switch to realizations consisting of a fourfold covered mesh of two triangles.

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Figure 3. This polyhedron called “Vierhorn” is locally rigid, but snaps between its spatial shape and two flat realizations; Below: Development of the “Vierhorn”; dashes indicate valley folds.

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It turns out that the computation of the spatial shape of any four-sided doublepyramid from given internal metric, i.e., from its 12 edge lengths, is an algebraic problem of degree 8. Hence, there are either up to 8 different realizations, or the edge lengths admit a flexible form. One needs to be careful when any polygonal mesh looks like a flexible one. A famous example is described in C. Schwabe’s and W. Wunderlich’s article [12] on a polyhedron exposed at the science exposition “Ph¨anomena” 1984 in Z¨ urich (see Fig. 3). At that time it was falsely stated that this polyhedron is flexible, but it is only snapping between two different flat realizations and one spatial shape. The development reveals that all faces of this polyhedron are congruent isoceless triangles. There are four mile-stones in the theory of flexible polyhedra: • The first important result in the theory of rigidity claims that every convex polyhedron is rigid [2]. This is due to A. L. Cauchy 1813. The example presented in Fig. 2 is no contradiction since the convex form, the regular octahedron, is locally rigid. • 1897 R. Bricard [1] classified all flexible octahedra, i.e., all flexible foursided double-pyramids with a not necessarily coplanar equator. However, all these polyhedra have self-intersections. A real-world model can only be built either as a wireframe or as a cardboard polyhedron where two faces are omitted. • R. Connelly detected 1977 the first flexible sphere-homeomorphic polyhedron without self-intersections and without twofold covered faces [3]. A simplified flexing sphere with 9 vertices was presented 1980 by K. Steffen [10] (compare [5, p. 347]) as a compound of two Bricard’s polyhedra. • 1996 I. Sabitov [6] proved the famous Bellows Conjecture stating that for every flexible polyhedron in E3 the oriented volume keeps constant during the self-motion [6]. This was a consequence of his generalization of Heron’s formula: For any orientable polyhedron with triangular faces in E3 there exists a polynomial whose coefficients are polynomials in the squared edgelengths over Q and which has the square V 2 of the volume as a root; this polynomial depends only on the combinatorial structure of the polyhedron. There is only an algorithm available for determining this polynomial. If Sabitov’s result had been known at the exposition of “Vierhorn” (Fig. 3), then it would have been evident that this is not really flexible because the volume changes drastically during the transition from the spatial form to the flat pose. There is still another kind of flexibility placed between riditity and continuous flexibility: According to W. Whiteley’s principle of averaging (see [11]) it can be seen as a limit of snapping polyhedra. In the following definition we make use of a standard notation of kinematics (see, e.g., [8]): Each infinitesimal rotation1 about an oriented axis with normalized dual coordinate vector b a ∈ R6 and signed angular b. The twist of the composition velocity ω can be represented by the twist vector ω a of two infinitesimal rotations is the sum of the two twist vectors. Definition 4. Suppose that to each internal edge eij of a polygonal mesh we can assign an angular velocity ωij for the relative motion of the adjacent fi against fj 1This is an affine map which appoints to each point in E3 a velocity vector.

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in such a way that for each loop f0 , . . . , fs = f0 of faces, where any two consecutive faces share an edge ei−1 i , i = 1, . . . , s , the sum of corresponding twists vanishes, Ps b . Then the mesh is called “infinitesimally flexible”. i.e., i=1 ωi−1 i b ei−1 i = 0 When the only compatible appointment of angular velocities to all internal edges is the trivial one with ωij = 0, then the mesh is called infinitesimally rigid. A real-word model of an infinitesimally flexible mesh shows an apparent, but somehow confined flexibility. At the Vierhorn both flat positions are infinitesimally flexible. A classical extension of Cauchy’s result states that each convex polytope is even infinitesimally rigid. Remark 1. a) Infinitesimal flexibility is usually defined for frameworks, but this works only for triangular meshes. Before applying it, e.g., to quadrangular meshes, we have to split all quadrangles by a diagonal into two triangles and to build pyramids over them in order to guarantee the planarity of the quadrangle. b) There are even different orders of infinitesimal rigidity to distinguish (e.g., [7]). But for the sake of brevity we focus here only on first-order infinitesimal flexibility. 2. Flexible polygonal meshes We are now concentrating on quadrangular meshes and start with a Kokotsakis mesh (German: Vierflach), the compound of 3 × 3 planar quadrangles, which is named after A. Kokotsakis [4]. In Fig. 4, left, the scheme of a quadrangular Kokotsakis mesh is shown with a central face f0 and a belt of 8 quadrangles around it. On the right hand side a flexion is displayed.

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Figure 4. Left: Scheme of a quadrangular Kokotsakis mesh. Right: Flexion of a flexible version; dashes indicate valley folds. A complete classification of all continuously flexible Kokotsakis meshes is still open (compare, e.g., [9]). However, the geometric characterization of infinitesimally flexible meshes has already been given in [4] (see Fig. 5). We follow Kokotsakis’ ideas and use in our proof of this characterization standard results from Kinematics (e.g., [8]): For any two faces fi , fj sharing an edge, this edge — here denoted by ij — is the axis of the relative motion. Due to the Three-Pole-Theorem for any three faces fi , fj , fk with rotations as pairwise relative motions the three relative axes ij, ik and jk must be coplanar and share a point, which is denoted by ijk.

ON THE RIGIDITY OF POLYGONAL MESHES

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The angles αj , αk which jk enclosed with ij and ik, respectively, define the ratio of angular velocities of fj and fk against fi by ωji : ωki = sin αk : sin αj . This implies, e.g., that at the Kokotsakis mesh (see Fig. 5) the axis 12 of the relative motion between f1 and f2 is the line of intersection between the planes spanned by f5 and f0 . We call the line of intersection between the planes of fi and f0 the trace of fi , i = 1, . . . , 8 . )

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Figure 5. Infinitesimally flexible Kokotsakis mesh; the bounding faces f1 , . . . , f8 are cut by a plane parallel to that of f0 . Theorem 5 (Kokotsakis, 1932). A Kokotsakis mesh is infinitesimally flexible if and only if the following three points are collinear: the points of intersection 013, 123 and 134 between the traces of the pairs of faces (f1 , f3 ), (f5 , f6 ) and (f7 , f8 ), respectively. This is equivalent to the statement that the points of intersection 024, 234 and 124 between the traces of the pairs of faces (f2 , f4 ), (f6 , f7 ) and (f5 , f8 ), resp., are aligned. Proof. Due to the Three-Pole-Theorem the traces 12 of f5 and 23 of f6 meet at point 123; the traces 14 of f8 and 34 of f7 meet at point 134; the traces 01 of f1 and 03 of f3 meet at point 013. In the infinitesimal case the three points 013, 123 and 134 must be located on the relative axis 13. Also by Desargues’ Theorem we can conclude that the collinearity of these three points is equivalent to the collinearity of points 024, 124, and 234. Conversely, the position of 13 defines the ratio of angular velocities of the faces f1 and f3 with respect to f0 . The other relative axes define the angular velocities of the other faces. Hence, the stated collinearity implies a compatible assignment of angular velocities to all internal edges.  An interesting continuously flexible quadrilateral mesh dates also back to Kokotsakis [4]. We start with a flat pose which consists of congruent quadrangles of a planar tessellation (Fig. 6). Any two quadrangles sharing a side (e.g., f3 and f4 ) change place under a rotation through 180◦ (= half-turn) about the midpoint (C)

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Figure 6. Kokotsakis’ flexible tessellation. of the common side. Any two adjacent quadrangles form a centrally symmetric hexagon, and the complete tesselation can also be generated by translations of this hexagon. The arrows in Fig. 6 indicate the directions of these translations. When the quadrangles are convex, then this polygonal mesh is flexible (Kokotsakis [4, p. 647]. We call this mesh a tesselation mesh, for short. In the following theorem we extend Kokotsakis’ result by characterizing the flexions of this polygonal mesh. We are interested on constrained motions of the mesh. Therefore we exclude additional degrees of freedom of single faces by the request: Whenever the tesselation mesh includes three faces with a common vertex, then also the fourth face of this pyramid must be included. On the other hand, when the basic quadrangle is a trapezoid, then there are aligned edges along which the mesh can be folded. We exclude these trivial flexes by requiring a generic basic quadrangle. Theorem 6. Let a polygonal mesh be extracted from the planar tesselation displayed in Fig. 6 in such a way, that with any three faces with a common vertex also the fourth face through this vertex is included. a) This quadrangular mesh is continuously flexible if and only if the initial quadrangle is convex. b) In the generic flexible case, at each non-planar pose of a continuous self-motion all vertices are located on a cylinder of revolution (Figs. 8 and 9). c) The faces of the flexion can be obtained from a line-symmetric hexagon composed from two adjacent quadrangles by applying iterated coaxial helical motions. In the flat pose these helical motions convert into the translations applied to a centrally symmetric hexagon in order to generate the planar tessellation. Proof. First we pick out the four faces f1 , . . . , f4 with the common vertex V1 (Fig. 7). These congruent faces form a four-sided pyramid which is flexible, provided the fundamental quadrangle is convex. Otherwise, one interior angle of a face at V1 would be greater than the sum of the other 3 interior angles so that the only realization is the flat pose. Let any non-planar flexion of this pyramid be given (Fig. 7, left). For any pair (f1 , f2 ), . . . , (f4 , f1 ) of adjacent faces there is a respective half-turn ρ1 , . . . , ρ4

ON THE RIGIDITY OF POLYGONAL MESHES

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f2 ρ1 (f4 ) ρ1 (f1 ) ρ2 f3 V2 ρ1 ρ3 f6 ρ1 ρ2 ρ1 f1 V 1 ρ1 (f3 ) ρ1 (f2 ) f4 ρ4 ρ (f ρ4 (f4 ) 4 1) e56 V3 ρ 4 ρ 3 ρ 4 V4 f5 ρ4 (f3 ) e58 ρ4 (f2 ) f8 f9

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Figure 7. The complete flexion can be generated by applying iterated half-turns ρi to an initial face f1 .

which swaps the two faces. So, e.g., f2 = ρ1 (f1 ) and f1 = ρ1 (f2 ). The axis of ρ1 (see Fig. 7, left) is perpendicular to the common edge V1 V2 , and it is located in a plane which bisects the dihedral angle between f1 and f2 . After applying all four half-turns ρ1 , . . . , ρ4 consecutively to the quadrangle f1 , this is mapped via f2 , f3 , and f4 onto itself; hence the product ρ4 . . . ρ1 equals the identity. (We indicate the composition of mappings by left multiplication.) Because of ρ−1 = ρi we obtain i (2.1)

ρ1 ρ2 = ρ4 ρ3 .

Lemma 7. The product of two half-turns about non-parallel axes a1 , a2 is a helical motion. Its axis is the common perpendicular of a1 and a2 ; its angle of rotation is twice the angle and the length of translation is twice the distance of the axes a1 , a2 . When our pyramid with apex V1 is not flat, then the axes of the half-turns are pairwise skew; the common perpendicular for any two of these axes is unique. Hence (2.1) implies that the axes of the four half-turns have a common perpendicular s. The motions ρ1 ρ2 = ρ4 ρ3 and ρ1 ρ4 = ρ2 ρ3 are helical motions with the common axis s. Now we extend the flexion of the pyramid with apex V1 stepwise to our polygonal mesh by adding congruent copies of the initial pyramid without restricting the flexibility: The rotation ρ1 exchanges not only f1 with f2 but maps the pyramid with apex V1 onto a congruent copy with apex V2 sharing two faces with its preimage. This is the area which is hatched in Fig. 7, right. Analogously, ρ4 generates a pyramid with apex V4 and sharing the faces f1 and f4 with the initial pyramid. Finally there are two ways to generate a pyramid with apex V3 . Either, we transform ρ2 by ρ1 and apply ρ1 ρ2 ρ1 , which exchanges ρ1 (f2 ) = f1 with ρ1 (f3 ) and swaps V2 and V3 . Or we proceed with ρ4 ρ3 ρ4 , which exchanges ρ4 (f4 ) = f1 with ρ4 (f3 ) and swaps V4 and V3 . Thus we obtain mappings (ρ1 ρ2 ρ1 )ρ1 = ρ1 ρ2 and (ρ4 ρ3 ρ4 )ρ4 = ρ4 ρ3 with f1 7→ f5 and V1 7→ V3 . Both displacements are equal by (2.1), and we notice (2.2)

ρ1 ρ2 = ρ4 ρ3 : f1 7→ f5 , f2 7→ ρ1 (f3 ), f3 7→ f1 , f4 7→ ρ4 (f3 ).

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Hence each flexion of the initial pyramid with apex V1 is compatible with a flexion of the complete 3 × 3 tesselation mesh. Is this the only flexion of this mesh induced by the given flex of the initial pyramid?

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Figure 8. For each flexion of the first and second kind the vertices are placed on a cylinder of rotation; the marked points are located on a helical line.

Lemma 8. A generic flexion of a 3 × 3 tesselation mesh is uniquely defined by the flexion of one included pyramid consisting of four faces with a common vertex V1 . Proof. At our previously defined flexion the pyramids with vertices V2 , V3 and V4 are congruent to that with vertex V1 . However, there is another possibility at V2 , which is compatible with that at V1 : We can reflect the two left faces f6 , f7 (Fig. 7, right) in the plane spanned by the edges e27 and e16 . In the same way it is possible at V4 to replace the faces f8 , f9 by their mirrors with respect to the plane e18 ∨ e49 . In total this gives 3 alternatives: Reflect either one pair of faces or both simultaneously. Does any of these alternatives keep place for inserting the last face f5 with given interior angle at V3 ? (i) When only f6 and f7 are reflected, the edge e56 is replaced by its mirror in the plane e27 ∨ e16 . This preserves the angle with e58 only if the plane e27 ∨ e16 passes through e58 . After applying ρ1 , this is equivalent to the statement that ρ2 (e14 ) is placed in e23 ∨ e14 , which means that the diagonal plane e23 ∨ e14 of our initial pyramid is the exterior bisector of the faces f2 and f3 . (ii) When reflecting f8 and f9 only, then the mirror plane e18 ∨ e49 must pass through e56 in order to preserve the interior angle of f5 at V3 . After applying ρ3 , this is equivalent to the condition that e12 ∨ e43 is the exterior bisector of the faces f3 and f4 . (iii) When e56 is reflected in e27 ∨ e16 and e58 is reflected in e18 ∨ e49 , then the angle between e56 and e58 is preserved if and only if there is a rotation mapping e56 and e58 onto their respective mirror images. The axis a if this rotation is the line of intersection between the two mirror planes. In the case of a rotation the axis a must span planes with e56 and e58 which enclose with e27 ∨ e16 and e18 ∨ e49 , respectively, the same oriented angle.

ON THE RIGIDITY OF POLYGONAL MESHES

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Only in particular poses condition (i), (ii) or (iii) can be fulfilled, but we excluded this by the request for a generic pose. Hence the flexion of the 3 × 3 tesselation mesh is uniquely defined.  By Lemma 8 we can uniquely extend the flexion of the initial pyramid to the 3 × 3 tesselation mesh and furtheron to the complete polygonal mesh, apart from particular poses. But under a continuous self-motion it is not possible to switch into one of the exceptional reflected positions listed above in (i), (ii) or (iii). Hence this mesh is continuously flexible, and all included four-sided pyramids are congruent. We detect at the flexion also the spatial analogues of the translations in the plane: The product ρ1 ρ4 = ρ2 ρ3 maps the pyramid with apex V1 onto that with apex V2 . On the other hand we have (2.3)

ρ1 ρ4 : f1 7→ ρ1 (f4 ), f4 7→ f2 , ρ4 (f2 ) 7→ f1 , ρ4 (f3 ) 7→ ρ1 (f3 ).

When f3 and f4 are glued together, we obtain a skew hexagon, one half of our initial pyramid with apex V1 . The half-turn ρ4 maps this hexagon onto itself; hence it is line-symmetric. By (2.2) the helical motion ρ1 ρ2 maps this hexagon onto the compound of f1 and ρ4 (f3 ) and furthermore f1 onto f5 . The inverse ρ2 ρ1 is the spatial analogon of the translation indicated in Fig. 6 by the red arrow pointing upwards to the right. On the other hand, ρ4 ρ1 maps the compound of f1 and ρ4 (f3 ) onto ρ1 (f4 ) and ρ1 (f3 ). It these two helical motions act repeatedly on the line-symmetric hexagon, the complete flexion is obtained. Since all vertices of the flexion arise from V1 by motions which keep the common perpendicular s of the half-turn axes fixed, e.g., V2 = ρ1 (V1 ), V3 = ρ1 ρ2 (V1 ), V4 = ρ4 (V1 ), they all have the same distance to s, i.e., they are located on a cylinder of revolution with axis s.  Remark 2. a) When starting from the flat initial pose of the pyramid with apex V1 , there are two self-motions possible since there are two edges of the pyramid where the adjacent interior angles at V1 have a sum smaller than 180◦ . These edges become valley-folds in Fig. 8. Hence our polygonal mesh admits two kinds of differentiable self-motions. Figure 9 shows snapshots of these two self-motions. b) In the case of a trapezoid f1 one kind of generating motion is a rotation about s, the other a translation along s. However, in this case we have a higher degree of freedom since the mesh can be bended along each serie of aligned edges. There is also a direct way to get a flexion of a tesselation mesh: We can start with any point V1 and with three half-turns ρ1 , ρ2 , ρ3 such that the axes have a common perpendicular. However, the quadrangle V1 . . . V4 will be planar only if V1 obeys an additional condition. Conclusion Not each polyhedron is uniquely defined by its development. We shed light on the different types of rigidity of polygonal meshes und characterize the flexions of Kokotsakis’ tesselation meshes.

HELLMUTH STACHEL

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k t l e i K o * * s l a a k la i s t e s s k i **Ko k o e ti k l o o t o o s a l k t a o i t K k k o k t o K o t a koo tssats o eeelll *KKo iisosooninn ottsk o skk a Kk **KK o *KK ak ak akk ataa** ottsteasssskk s iaeesaslsl* **t*e*ssttsookooK * oo kiisis *Koko a s s o k o a * s * tsakis* k * o s e k t t k s K o o s tessell o s t t o K ation* * * k t a s e s o Kokot e sakis* k * K k o *nnK ssssss *K okkoo lliilislsssaeette* o t*ikti*oo*nss*ii*K okk ottsaask K * nK*oK ookkoootstatsksaiakik*sitk*ikkesaassse*sssetlloollaakkttiiooonK s a o t o o k o o a * i k i nK**Kok ot o s K k * o t t t o s s K o t a t o s ttteeeess ***K * o s a o o e is k nn*KK***K e k K nK soiaitkktlsssaalleaasalek * siistsssio* i*asssst*eik o*nnK **so nn***K KKoookkkoots saakkii o ttiiiio* K oookoikt kotkosoetttsttks** Ko isksieaalssls*lattookkooK**Kok ootksakiss*akkis* s*** tt s lek ess*n *K * ok n*K io * * t t o t k s t s K a o a * l i * t l a a i a n i o k * a s o k s s s l a i a k o s o o * K l t t i k * k o e t * k o t * a o o a k t o K n i t K i s * K s o s k K i o k i k o i a * s K n i kkio k kkiiiiisiss *K o o o * lloltelee s i l k K o * i k * o K kk otstssaakkiiisiss****tteetlelllslllalaaatttiiiio * s t oo n* so*ttsa * o i * elkeassksssksssssskieeoeisstt**t** nnnooo*oitK ititttatttaassassaaatok Kno aaK sssaikkaisst* s iioo k ooknno*K n o K *n* *K o n o o k llltllselllloaeo a iss** stsseeesessslslellaattio ooo K n ssaaaaaakkkk **Kokootts aatK K on* ott* * * *onnoonkiii* nnik staokkis ** ats *K llsesekossikkeoto**tsssiikakaksiitsso* ioookkot Kook akis K ko saakkis Ko tta *osn a salK * stes s elllat n* e k a k o s k k * i e * s l s* l o * s o o te s s a k ootttttststssssaa *K ki * l e t ts o l K i i n l K l n sa s n o t s s t s ak * K ot t s * a a e K a o k s a t a o l i n K t ok * is l l e o a s * o K * ok * * * K K t e ots t k ti i n* e e te K o s o t s a e aki n a t s t s tio n s a s o k l k s*t o ll * ti i t i * ess o el s l i t i l * o e a t ell i t la k ss e o s t t * * t l on t te *K ki 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*sikaassttookkooK t o o k k l K s * t l k o K k e l o i a e t * o o t a t k l a s t t a i s t k l on kikassstto okos K sss ellla oel llaal tiioon toookkoosK tio kaasassttosoakkoosK K sn* ell*lnlaootiiititaao n oita tes* s*s*siiiskk o t k *sik t o i * k s l t * e s * t i n t * k a s e s K * o K a k a * e i a o on akis K s to koKise llola ti n* oK *K o K * ko K*n ll s ikk aa sKtooktoo e ki olKl*aKla**tntniiooion *K ll*n atti on K K asto K *ll llas* tio n**K * kokoK o tK o n *K ookkototastokkooK*K* oask ita ikaass*sitttko K k a ioo * ookoK ok n t o o *sik stoko a o k * * * n t s i i s K k t o* io nn* K ko tsa ti o *K ok ot n s*t to ikkkaoosoKtok* o **K o ***soiK a a o k * k t k * n i s o n s K i * o k * K K o o K *s n K ko tsa kis*n *K ok ots aki aKs * *ksko ikK Koooakkkais*ts* *tK * ok o* * K * * Ko ko K o o*K sao K sK ** ki oost*s o*sk K k 1 *Ko 2 o *K K ok tso ko ko tsa kis* *K ok ots akiss* o o o k k t akkkk ottts * tssa kio a tsak a s o k * k ak a s * s o k o i s a i a is*t i o * ts * *tttteeeessssseklolat tsaak is* kkiis * kiisssss* k esse * is ts *s a o k t i e s llati l t o s s t a * * l e ak n* k *iK * n on ssseeelllllaaasattttiiio t s*ok on* is K *isK ok ottsak sa n* on ko Kok K o *o is llaattiiiko o o * o * s k i t a s k s o o * * k i K t a k n t * o tsak i k o s * s o i s * * a i K * k t s * i n o o * k o s * K is** i a * o k * K s s * t K k n * * o * o s K K kottsa akiis* *K * *K K Kook Kooookk k a **KKKK otsssa **K kiiisss* K **K a K oookk * o K K * k * o o o o k * t o * a k k * o s K o o s 4 Ko K 3 * k * K k k * o k s o s i * o k t K o k k a o o ok K*oKk k oookoooookoottttts otsssaakkkiiss*** o k o ottttttssssssaaaakkiiss*te s ak ko k saak kottottsssssttstsaasaaaaakkkkkkiiiiiissss****ttttteeeesssssseeellllaattiioon*Kokotts aakkkiisss** kkiiiiiissss****ttteeessssssseeelllllaattion* k ss* is *ttessselllllation k i s Koko *Kok s**ttteessssseeelllllaattiiioonn*K * i s* s i a n s i a s t * e o k i s t * e t s Kokotsakis* ss***tteessseellllaatiionn**K okot akis* i Ko eesssseeelllllaaatttiiooonn***K K * t k s s a o a l o o s e a o n k t e s k ko sakiis* ok tessssseeelllllaaatttiiiooonn***K Kok ottsa is* kotts Ko ellllaaattiiioooonnn***K K s* ak s sak ko ko kiis* i ok o a K t t n o o * s e l k i t o K i a t l o t attiioonnn**K a o s * * n l o s* Ko ko llaatt ionn**K o aki s* k kottssa ak Ko K k kis ok k * akis* okotttssa otttssa onn**K ak tiiiooonn **K o Ko * Kooo kiiiisss** o o otssa * kis* k k a * o ak t * * o * k s o a * K * s k * k K t o * * * * n n o * t * i KKKoo o a K K s *K K s ***K s K K * kiisss* t k a * * o K k K K K K o K * t i o * o * s k a * * * s t k i K * K t k o K a K s k * o k * k * o a s o o s o i K * t o o * o K K i o * o * * K t s k o a s o k K a t * K K o K k o 5 * 6 k s k o o * k * tsakis* k t k K s k k s o * n*Koko o i o oo kkoooooootttsssakis*te k ssellatio *K * **K ok k ookkkkooktoosotttssaaakkiiiss*** *Kokot ooookk o * K**K o ootsakis*tessel ion*Kkok kkkokkokkkokoots K oottsttots o akis ssaaakkkkak **K k oootttsstaataskskia kkisss* iisssis* K kokokko K oottttts **ooK t s* kiiis oookkooototttttttstsssststsssaaaaaaaakkkkiiiissss****ttteeessssseellllat i s s sssaaa *ttte K t s ation* k * s * i s s e t a a e k K tssaskkaiisskk*iiss** * s ll s k s k i i e *te ss ation* se se iss* ss ation Kokots ookoottssaaaakk ellla is* k elllla k s k es *ttte i * i * K e K a k * i *K o s k l l kkkissisi***ikssis * i i t i e K t koaa t is a a i s * Kookkooottsssaaaakkkkkiiiiisssss*** * i s s i s k s on* okots s** * s on t t ts e i o k * e s s i i s a n t o o t l s e * e k s s *** i a t l * * akis**Koko Koakis* ia Ko *K is*tttteeeeesssssellllatio n*K tsakkiiiss**tteeesssssseeellllaaattio * * o kot ss s * k * t t o o n s o a o l k k k k e a o e * n n *Kokots l io sak ots s*tesssell sa Kok otts on* *K no* *K Kn ak ltltliiaaotttnii**Kok *oK kis**Kokotsakiis*te ssssssseeseeellllelllaaation otsak a i * is* akis*tteesssssssseeeeeelllllllllalllltaaaattttiiio okkis* K ok otsa is**Koko n* K iaioon*Kokotsakis* ookotsa tsakis**tessella *kkK *Kokotsakis*tessellati otsa kiss* nniio se lllllllaalllttion*K oak n o o * s i a o t ellllaaaaattiiitooiinoon K t o e t t s k o * o n K k * t k s i o n i t a o o t * kis* s o i n s K K k t s a o s * * a i i a a t ttiitioooionno***nnnK o * s * a n k k * t t o t K o o s i o k o K k s k n t s*ak is**Kok ot*s ak kK ioonno**nnK K t t o kkkooooots i o K kaao o k K o * i i is K o t **oooK nnnn**K t o s o t K s i s t k a * k a K * s k a o s nnn**K koooot kkkoitsts* kiiss* K*oK **KoKkkoooottks ooktitss*aakki *

ON THE RIGIDITY OF POLYGONAL MESHES

17

Acknowledgment. This research is partly supported by Grant No. I 408-N13 of the Austrian Science Fund FWF within the project “Flexible polyhedra and frameworks in different spaces”, an international cooperation between FWF and RFBR, the Russian Foundation for Basic Research. References [1] R. Bricard: M´ emoire sur la th´ eorie de l’octa´ edre articul´ e. J. math. pur. appl., Liouville 3, 113–148 (1897). [2] A.L. Cauchy: Recherche sur les poly` edres – second m´ emoire. J. Ecole Polyt´ echnique 9, 87–96 (1813). [3] R. Connelly: A Flexible Sphere. Math. Intell. 1, no. 3, 130–131 (1978). ¨ [4] A. Kokotsakis: Uber bewegliche Polyeder. Math. Ann. 107, 627–647 (1932). [5] E.D. Demaine, J. O’Rourke: Geometric folding algorithms: linkages, origami, polyhedra. Cambridge University Press 2007. [6] I.Kh. Sabitov: The volume of a polyhedron as a function of its metric. Fundam. Prikl. Mat. 2, 1235–1246 (1996). [7] H. Stachel: Higher-Order Flexibility for a Bipartite Planar Framework. In A. Kecskem´ ethy, M. Schneider, C. Woernle (eds.): Advances in Multi-body Systems and Mechatronics. Inst. f. Mechanik und Getriebelehre, TU Graz, Duisburg 1999, 345–357. [8] H. Stachel: Teaching Spatial Kinematics for Engineers. Proceedings ICEE 2005, Gliwice/Poland, (ISSN 1562-3580), vol. 2, 845–851. [9] H. Stachel: A kinematic approach to Kokotsakis meshes. Comput. Aided Geom. Des. 27, 428–437 (2010). [10] K. Steffen, A symmetric flexible Connelly sphere with only nine vertices. http://www.math. cornell.edu/~connelly/Steffen.pdf [11] W. Whiteley, Rigidity and scene analysis. In J.E. Goodman, J. O’Rourke (eds.): Handbook of Discrete and Computational Geometry, CRC Press, Boca Raton, New York 1997. [12] W. Wunderlich, C. Schwabe: Eine Familie von geschlossenen gleichfl¨ achigen Polyedern, die fast beweglich sind. Elem. Math. 41 (1986), 88-98. Institute of Discrete Mathematics and Geometry, Vienna University of Technology, Vienna, Austria E-mail address: [email protected]