kokotsakis meshes and ﬂexible quad meshes · 2013. 7. 2. · any plane quadrangle is a tile for a...

Kokotsakis meshes and

flexible quad meshes

Hellmuth Stachel

[email protected] — http://www.geometrie.tuwien.ac.at/stachel

Joint Internat. Mathematics Meeting, University of Alba Iulia /Romania, June 27–30, 2013

Special Session: Articulated Systems: Combinatorics, Geometry and Kinematics

Outline

1. Flexible Kokotsakis meshes

2. Kokotsakis’ flexible tesselation

3. Rigidity of a quadrangular cylinder tiling

Funding source: Grant No. I 408-N13 of the Austrian Science Fund FWF

within the project “Flexible polyhedra and frameworks in different spaces”, an

internat. cooperation between FWF and RFBR (Russian Foundation for Basic Research)

Acknowledgement: Georg Nawratil, TU Wien

June 29, 2013: Articulated Systems: Combinatorics, Geometry and Kinematics, University of Alba Iulia / Romania 1/36


V1

V2

V3

V4

f0

f1

f2

f3

f4

a1

a2

a3

a4

Special case n = 4

A Kokotsakis mesh is a polyhedralstructure consisting of an n-sidedcentral polygon f0 surrounded by abelt of polygons.

Each side ai , i = 1, . . . , n, of f0is shared by a polygon fi . At eachvertex Vi of f0 four faces are meeting.

Each face is a rigid body; onlythe dihedral angles can vary (‘rigidorigami’). Can it be continuouslyflexible ?



V1 V2

V3

W3

W1

W2

f0

f1

f2f3

a1

a2

a3

Special case: n = 3

A Kokotsakis mesh for n = 4 is alsocalled Neunflach [German] (nine-flat)

(Kokotsakis 1931, Sauer 1932)

For n = 3 the Kokotsakis mesh isequivalent to an octahedron withV1V2V3 and W1W2W3 as oppositetriangular faces.

This offers an alternative approach toR. Bricard’s flexible octahedra.


G. Nawratil: There is exactly one flexible octahedron with vertices at infinity.

α1

α2

a

b

c

a′

b′

c′

It is free of self-intersections !

a

b

c

a′

b′c′

. . . a particular T-flat



In discrete differential geometry there is ainterest on polyhedral structures composedof quadrilaterals (quadrilateral surfaces).

If all quadrilaterals are planar, they form adiscrete conjugate net = quad mesh. H. Pottmann, Y. Liu, J. Wallner,

A. Bobenko, W. Wang:

Geometry of Multi-layer Freeform

Structures for Architecture. ACM Trans.

Graphics 26 (3) (2007), SIGGRAPH 2007



In ‘Freeform Architecture’ most of the surfaces are designed as polyhedral surfaces

— like the Capital Gate in Dubai (height=160 m, inclination 18◦)Steel construction: Wagner Biro, Austria



Theorem: [Bobenko, Hoffmann,Schief 2008]

A discrete conjugate net in general position

is continuously flexible ⇐⇒ all its 3 × 3complexes are continuously flexible.

The classification of all flexible Kokotsakismeshes (n = 4) has recently be finished:

A historical model of a flexibleKokotsakis meshes:

Ivan Izmestiev: Classification of flexible Kokotsakis polyhedra with quadrangularbase, preprint, 74 p., May 2013



V3

V4

f0

f1

f2

f3

I10

I20

I30

α1

β1

γ1

δ1

α2

β2 γ2

δ2

Transmission from f1 to f3via V1 and V2.

I10I20

I30

A1B1

A2

B2

ϕ1ϕ2

ϕ3

α1 β1

γ1

δ1

α2β2

γ2

δ2

The composition of two spherical four-bars.



Continuous flexibility of a Kokotsakismesh for n = 4 means:

The transmission from f1 to f3 can bedecomposed in two different ways, viaV1 and V2 – or via V4 and V3.

V1

V2V3

V4

f0

f1

f2

f3

f4

I10

I20

I30

I40



Conditions for one flexible case:

α1 + β2 = δ1 + δ2

sα1sγ1 : sβ2sγ2 = sβ1sδ1 : sα2sδ2 =(cβ1cδ1 − cα1cγ1) : (cβ2cγ2 − cα2cδ2)

Right: The spherical image of thiscase with two decompositions.

A second pose is shown in light-blue.

I10

ϕ1

I30

ϕ3

I20

A1B1

A2

B 2

ϕ2

α1β1

γ1

δ1

α2

β2

γ2

δ2

I ′20

A ′1

B′

1

A′2

B′

2

ϕ′2

α ′1

β ′1

γ ′1

δ′1

α′2

β′

2

γ′

2

δ′2

ψ

ψ

ψ



Only in very particular cases we know something about the geometry of theflexions obtained during the self-motions of continously flexible Kokotsakis meshesand quad meshes,

e.g., application of a discrete Voss surface in

Nadja Posselt: Synthese von zwangläufig beweglichen 9-gliedrigen Vierecksflachen,diploma thesis, TU Dresden 2010



Cf3

f4 f2

f1

A. Kokotsakis, 1932Athens

Any plane quadrangle is a tile fora regular tessellation of the plane(wallpaper group p2, generically).

It is obtained by applying

• iterated 180◦-rotations about themidpoints of the sides of an initialquadrangle or

• by applying iterated translationson a centrally symmetric hexagon.

For a convex f1 this polyhedralstructure is continuously flexible.



f4

f3 f2

f1r

l

A. Kokotsakis, 1932Athens

Any plane quadrangle is a tile for aregular tessellation of the plane.

It is obtained by applying

• iterated 180◦-rotations about themidpoints C of the sides of aninitial quadrangle or

• by applying iterated translationsl, r on a centrally symmetrichexagon.

For a convex f1 this polyhedralstructure is continuously flexible.



V1

V2

V3 V4

f1

f2 f3

f4

f7

f6

f5f8

f9

We pick out a 3 × 3 complexfrom this tessellation. Why it isflexible ?

We first focus on the four-sidedpyramide with vertex V1.

Due to the required convexity nointerior angle is > π . Thereforethis pyramid is continuouslyflexible.



f1

f2

f3

f4

V1

V2

V3

V4

ρ1ρ2

ρ3

ρ4

In each pose for any twoneighbouring faces there is a180◦-rotation ρi which inter-changes these two faces.

The axis is located in a bisectorplane and passes through themidpoint of the common edge.



The product of any two 180◦-rotations ρ1, ρ2about respective axes l1 and l2 is a helical motionρ2 ◦ ρ1 about the common perpendicular p :

angle of rotation: 2ϕ

length of translation: 2d l1

l2p

ϕ

d



V1

V2

V3 V4

ρ1ρ2

ρ3

ρ4

f1ρ1(f2)ρ4(f4)

f2ρ1(f1) f3

f4ρ4(f1)

After applying all four 180◦-rotations consecutively to thequadrangle f1, this is mappedvia f2, f3, f4 onto itself, hence

ρ3 ◦ ρ4 = ρ2 ◦ ρ1 .

=⇒ the four axes have a common perpendicular p, and all vertices V1, . . . , V4are at the same distance from p.



V1

V2

V3 V4

ρ1ρ2

ρ3

ρ4

f1ρ1(f2)ρ4(f4)

f2ρ1(f1) f3

f4ρ4(f1)

f7=ρ1(f4)

f6=ρ1(f3)

f5 f8 f9

ρ1 maps the pyramide withvertex V1 onto the pyramidewith vertex V2.

(There would also be a secondpossibility to continue theflection of the 2 × 2 mesh tof6 and f7.)

There are two possibilities to continue the flection onto the fourth pyramide withvertex V3 .



V1

V2

V3 V4

ρ1ρ2

ρ3

ρ4

f1ρ1(f2)ρ4(f4)

f2ρ1(f1) f3

f4ρ4(f1)

ρ1(f4)

ρ1(f3)

f5 ρ4(f3) ρ4(f2)



There are two possibilities to continue the flection onto the fourth pyramide withvertex V3 .



V1

V2

V3 V4

ρ1

ρ2

ρ3

ρ4

f1ρ1(f2)ρ4(f4)

f2ρ1(f1) f3

f4ρ4(f1)

ρ1(f4)

ρ1(f3)

f5 ρ4(f3) ρ4(f2)

ρ4ρ3ρ4

ρ1ρ2ρ1

ρ1 ◦ ρ2 ◦ ρ1 maps the pyramidewith vertex V2 onto the pyramidewith vertex V3.

ρ4 ◦ ρ3 ◦ ρ4 maps the pyramidewith vertex V4 onto the pyramidewith vertex V3.

f5 is the image of f2 = ρ1(f1)under ρ1 ◦ ρ2 ◦ ρ1, and image off4 = ρ4(f1) under ρ4 ◦ρ3 ◦ρ4, asf5 = ρ1 ◦ ρ2(f1) = ρ4 ◦ ρ3(f1).

The complete pose arises from the “hexagon” f3∪f4 under iterations of the helicalmotions r = ρ2 ◦ ρ1 and l = ρ4 ◦ ρ1 . This can be continued to a m× n mesh.



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At each flexion obtainable by a continuous self-motion of an m× n tesselationmesh all vertices are located on a right circular cylinder(discrete conjugate quadrangular net on this cylinder)



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Each flexion of an ∞×∞ tesselation mesh is periodic. Generically, its group ofisomorphisms is generated by coxial helical motions l, r and by ρ4. It is

isomorphic to that of the flat case (p2).



V1

V2

V3

V4

ρ2ρ1ρ2

ρ2

ρ3ρ3ρ4ρ3

p

e

Another method to obtain the flexions: The plane spanned by f11 intersects thecircumcylinder along an ellipse e.



V1

V2

V3

V4

ρ2ρ1ρ2

ρ2

ρ3ρ3ρ4ρ3

p

e

Conversely, there is a one-parameter set of ellipses passing through a given convexquadrangle. For each ellipse there are two cylinders of revolution passing through.Each such cylinder defines a flexion of a tessellation mesh (TM).



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7

For quadrangles with circumcircle the mesh (=Voss surface) admits a second flatpose with coinciding circumcircles coincide (not free of self-intersections !)


3. Flexion which tiles the cylinder


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Left: Flexions of a 9×6 TM

Under which conditionsthe flexion is horizontallyclosing, i.e., the right borderzig-zag fits exactly to theleft border — apart from avertical shift ?

(the trapezoidal case withaligned borders is excluded)



A m × n tessellation mesh is agrid of m × n quadrangles denotedby fij, 1 ≤ i ≤ m, 1 ≤ j ≤ n

fm1 fm2 fm3 · · · fmn... ... ... ...

f21 f22 f23 · · · f2nf11 f12 f13 · · · f1n

f22

f12

f23f32

f21

f13

f33

f31

f11

V1

V2

V3

V4



r1

r4

ρ2

ρ3

f11

ρ4

l−1r

l−1rρ4

l−2r2

l−2r2ρ4

l−3r3

l−3r3ρ4

lr

lrρ4

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l−1r3

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l−2r4

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l2r2

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lr3

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r4

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l−1r5

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l3r3 l2r4 lr5 r6

l−1ρ4 l−2rρ4 l

−3r2ρ4

r

rρ4

l−1r2

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l−2r3

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lr2

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r3

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l−1r4

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l2r3

l2r3ρ4

lr4

lr4ρ4

r5

r5ρ4

This scheme of a 7 × 7 TM shows whichproduct of l, r and ρ4 must be applied to f11to obtain fij.

Each flexion of an ∞ × ∞TM is periodic; the group ofisomorphism acts transitivelyon the quadrangles.

Lemma: f11 can be trans-formed into fij according to

fij =

li−j2 r

i+j2 −1(f11)

for i+ j ≡ 0 (mod 2),

li−j−1

2 ri+j−3

2 ρ4(f11)

for i+ j ≡ 1 (mod 2).

(r = ρ2 ◦ ρ1 and l = ρ4 ◦ ρ1)



r1

r4

ρ2

ρ3

f11

ρ4

l−1r

l−1rρ4

l−2r2

l−2r2ρ4

l−3r3

l−3r3ρ4

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lr3

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r4

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l−1r5

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l3r3 l2r4 lr5 r6

l−1ρ4 l−2rρ4 l

−3r2ρ4

r

rρ4

l−1r2

l−1r2ρ4

l−2r3

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r3

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l−1r4

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l2r3

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lr4

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r5

r5ρ4

For odd m f2n+1 = r(f1n) is identical withfk+2 1 of the most-left row, k ≡ 1 (mod 2), iff

l1−n2 r

n+12 = l

k+12 r

k+12 d(2π)

Since the involved helicalmotions commute pairwise, weobtain

l−n+k2 ◦ r

n−k2 = d(2π).

Theorem:

A flexion of an m×n TM closeshorizontally with vertical shift k⇐⇒

∃ a, b ∈ Z : larb = d(2π).

n = −a+ b, shift k = −a− b.



Example: Horizontally closing 7× 9 TM satisfying l−6r = d(2π):

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f 91

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f 71

f 61

f 51

f 41

f 31 f 21 f11

f 92

f 82

f 72 f62 f 13

f 24

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f2

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f27 f17

How to obtain a closingversion ?

• Either numerically byminimizing a certaindistance.

• Or by starting with twocoaxial helical motionsr, l obeying larb =d(2π) and ρ4.



After specifying ρ1, ρ2, ρ4 and any first vertex V1 a quadrangle V1 . . . V4 is defined.However, this will not be planar.

Lemma:

For planarity of the quadrangle V1 . . . V4 it is necessary and sufficient to specify V1on a ruled surface of degree 3.



A basic trapezoid yields as nontrivial closing flexion of a 16 × 17 TM a modifiedSchwarz lantern satisfying l−10r6 = d2π and k = 4.

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f 115

f 1616

f 1416

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f 1016

f 816

f 616

f 416

f 216



Theorem: For given axes of the 180◦-rotations ρ1, ρ2, ρ4 there is a two-parameterset of planar convex quadrangles for tiling a cylinder.


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f118

18× 3 TM with flexion

satisfying l−3r6 = d(2π)(bandage)


Rigidity of a quadrangular cylinder tiling

Theorem:

When at a horizontally closing flexion of an m × n TM the right and theleft border line are glued together — at least at one vertex — then in thecase of a non-cyclic base quadrangle the resulting quad mesh is infinitesimally rigid.

Remark: For a cyclic quadrangle V1 . . . V4 the clos

kokotsakis meshes and ﬂexible quad meshes · 2013. 7. 2. · any plane quadrangle is a tile for a...

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