building with butterflies: folding concentric pleated polygons

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Building with Butterflies: Folding Concentric Pleated Polygons Bryan Gin-ge Chen Leiden University Thursday, January 10, 13

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  • Building with Butterflies: Folding Concentric Pleated Polygons

    Bryan Gin-ge ChenLeiden University

    Thursday, January 10, 13

  • Origami is an artform aimed at creating 3D objects from 2D paper:

    Tomohiro Tachis Origamizer, 2011

    Satoshi Kamiya

    rigid origami

    Thursday, January 10, 13

  • Daniel Piker, after Ron Resch, Ben Parker and John Mckeeve

    http://spacesymmetrystructure.wordpress.com/2009/03/24/origami-electromagnetism/

    Origami as metamaterial

    Thursday, January 10, 13

    http://spacesymmetrystructure.wordpress.com/2009/03/24/origami-electromagnetism/http://spacesymmetrystructure.wordpress.com/2009/03/24/origami-electromagnetism/

  • Origami corrugations on flickr.comThursday, January 10, 13

  • The study of origami as a method of making new materials is still in its infancy.

    Origami artists have been mostly alchemists; we should attempt to develop the corresponding chemistry.

    Robert Langs Treemaker / Universal Molecule algorithm

    Bowers, Streinu, Proc. ADG 2012

    Thursday, January 10, 13

  • The butterflies of my first slide turn out to be a useful family of atoms for thinking about rigid origami.

    I will discuss mostly the geometry of the substance / molecule known as the hypar.

    Thursday, January 10, 13

  • erikdemaine.org/hypar

    Thursday, January 10, 13

  • erikdemaine.org/hypar

    Christmas tree star

    Thursday, January 10, 13

  • Demaine, Demaine, Lubiw 1999

    Polyhedral sculptures

    Thursday, January 10, 13

  • Fernando Sierra, 2009

    http://www.flickr.com/photos/elelvis/3179640715/in/photostream

    p- and d- surfaces

    Thursday, January 10, 13

    http://www.flickr.com/photos/elelvis/3179640715/in/photostreamhttp://www.flickr.com/photos/elelvis/3179640715/in/photostream

  • ????

    ...

    Thursday, January 10, 13

  • Demaine et al, 2011

    Key ideas:Interior polygons must remain planar.

    A vertex with all mountain or all valley folds coming out cannot be folded with planar faces.

    However, any folded internal ring would induce such a vertex by extending the paper.

    Thursday, January 10, 13

  • What actually happens?

    Thursday, January 10, 13

  • http://www.grasshopper3d.com/forum/topics/folding-paper

    Mrten Nettelbladt, 2010

    Thursday, January 10, 13

    http://www.grasshopper3d.com/forum/topics/folding-paperhttp://www.grasshopper3d.com/forum/topics/folding-paper

  • http://www.grasshopper3d.com/forum/topics/folding-paper

    Mrten Nettelbladt, 2010

    Thursday, January 10, 13

    http://www.grasshopper3d.com/forum/topics/folding-paperhttp://www.grasshopper3d.com/forum/topics/folding-paper

  • Demaine et al, 2011

    ????

    Non-alternating Alternating

    Thursday, January 10, 13

  • These two do exist for all angles and all layers! (But they might self-intersect)

    BGC 2011Thursday, January 10, 13

  • BGC 2011

    First, these crease patterns can be decomposed into butterflies:

    Thursday, January 10, 13

  • We build the hypar from the innermost layer outwards.

    The first stage is just a square folded by some angle along its diagonal.

    Can we glue the first butterfly?

    BGC 2011Thursday, January 10, 13

  • Each butterfly has a single degree of freedom which changes the distance between the wingtips.

    This distance is maximized (or minimized) when the butterfly is planar.

    If the distance between the tips of the target is in between these values, we can always glue the butterfly either above or below the plane through ABC.

    Can we glue the first butterfly?

    BGC 2011Thursday, January 10, 13

  • The distance shrinks!

    |~a+~b| < |~a|+ |~b|

    BGC 2011Thursday, January 10, 13

  • Distances always shrink!

    Therefore, these exist (as possibly self-intersecting immersions) for all angles and all numbers of layers!

    BGC 2011Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    How close are they to the real Hyperbolic Paraboloid?

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    x

    a

    yb

    =2z

    u

    x

    a

    +y

    b

    = u

    x

    a

    +y

    b

    =2z

    v

    x

    a

    yb

    = v

    The real Hyperbolic Paraboloid

    Doubly ruled:

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    x(u, v) = a

    u+ v

    2

    y(u, v) = b

    u v2

    z(u, v) =uv

    2

    The real Hyperbolic Paraboloid

    Parametric form:

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    The real Hyperbolic Paraboloid

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    The real Hyperbolic Paraboloid

    (0, bk,k2/2)(ak, 0, k2/2)

    (ak, 0, k2/2)

    (0,bk,k2/2)

    4-bar linkage

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    The real Hyperbolic Paraboloid

    (0, bk,k2/2)(ak, 0, k2/2)

    (ak, 0, k2/2)

    (0,bk,k2/2)

    4-bar linkage

    BGC 2011

    Fit the creases!

    Thursday, January 10, 13

  • x

    a

    2

    y

    b

    2= 2z

    The real Hyperbolic Paraboloid

    (0, bk,k2/2)(ak, 0, k2/2)

    (ak, 0, k2/2)

    (0,bk,k2/2)

    4-bar linkage

    +(0, 0,)

    BGC 2011

    Fit the creases!

    Thursday, January 10, 13

  • I computed the change in a, b, from layer to layer as an asymptotic series in 1/k...

    (0,bnkn,k2n/2 +n)(ankn, 0, k

    2n/2 +n)

    (0,bn+1kn+1,k2n+1/2 +n+1)(an+1kn+1, 0, k

    2n+1/2 +n+1)

    d(Vi,j , Vi,j1) = pi,j1

    d(Vi,j , Vi,j) = pi,j

    d(Vi,j , Vi,j+1) = pi,j1

    BGC 2011Thursday, January 10, 13

  • nonalternating:

    alternating:n+2 n

    =a2 + b2p

    2k2 a+ 2b

    2k3+O(k4)

    kn+2 k

    =

    p2

    k

    k3+O(k4)

    an+2 a

    =

    k3+

    a(b2 a2)p2k4

    +O(k5)

    bn+2 b

    =b(a2 b2)p

    2k4+

    (2a2 + ab+ 2b2)

    2k5+O(k6)

    n+2 n

    =p2(a2 b2) +

    2k2 3(a+ b)

    4k3+O(k4)

    kn+2 k

    =

    p2

    k

    k3+O(k4)

    an+2 a

    =

    2k3+

    p2a(b2 a2) + (b 2a)

    2k4+O(k5)

    bn+2 b

    = 2k3

    +

    p2b(b2 a2) + (a 2b)

    2k4+O(k5)

    Non-alternating Alternating

    (I dropped most subscripts when they are n) BGC 2011Thursday, January 10, 13

  • kn+2 k

    p2

    kkn

    q2p2n+ C

    Nonalternating:

    n+2 n

    p2(a2 b2) +

    2k2

    =O(1/n)

    an+2 a

    =O(n3/2)

    bn+2 b

    =O(n3/2)

    Non-alternating Alternating

    BGC 2011Thursday, January 10, 13

  • kn+2 k

    p2

    kkn

    q2p2n+ C

    Nonalternating:

    n+2 n

    p2(a2 b2) +

    2k2

    =O(1/n)

    an+2 a

    =O(n3/2)

    bn+2 b

    =O(n3/2)

    Non-alternating Alternating

    n = O(log n)

    BGC 2011Thursday, January 10, 13

  • kn+2 k

    p2

    kkn

    q2p2n+ C

    Nonalternating:

    n+2 n

    p2(a2 b2) +

    2k2

    =O(1/n)

    an+2 a

    =O(n3/2)

    bn+2 b

    =O(n3/2)

    Non-alternating Alternating

    n = O(log n)

    Folding the nonalternating pattern doesnt approximate a hyperbolic paraboloid! BGC 2011

    Thursday, January 10, 13

  • Non-alternating Alternating

    kn+2 k

    p2

    kkn

    q2p2n+ C

    Alternating:We can show that an bn = O(1/k)

    n+2 n

    =a2 + b2p

    2k2 a+ 2b

    2k3+O(k4)

    =O(k3)

    =O(n3/2)

    an+2 a

    =O(k3) = O(n3/2)

    bn+2 b

    =b(a2 b2)p

    2k4+O(k5)

    =O(k5) = O(n5/2)BGC 2011

    Thursday, January 10, 13

  • Non-alternating Alternating

    kn+2 k

    p2

    kkn

    q2p2n+ C

    Alternating:We can show that an bn = O(1/k)

    n+2 n

    =a2 + b2p

    2k2 a+ 2b

    2k3+O(k4)

    =O(k3)

    =O(n3/2)

    an+2 a

    =O(k3) = O(n3/2)

    bn+2 b

    =b(a2 b2)p

    2k4+O(k5)

    =O(k5) = O(n5/2)

    Everything converges! Whew!

    BGC 2011Thursday, January 10, 13

  • Non-alternating Alternating

    5000 10000 15000 20000 25000layer

    0.5

    1.0

    1.5

    2.0

    2.5

    3.0a,b

    alternating

    5000 10000 15000 20000 25000layer

    0.3

    0.4

    0.5

    0.6

    Dalternating Heven>oddL

    5000 10000 15000 20000 25000layer

    0.5

    1.0

    1.5

    2.0

    2.5

    3.0a,b

    nonalternating Ha>bL

    5000 10000 15000 20000 25000layer

    -0.4

    -0.2

    0.2

    0.4

    0.6

    Dnonalternating Heven>oddL

    Non-alternating

    Measurements:

    BGC 2011Thursday, January 10, 13

  • 0.5 1.0 1.5 2.0 2.5 3.0q

    -4

    -2

    2

    4

    6

    a,bAspect Ratio for nonalternating pattern

    =0.1

    =/2-.01BGC 2011

    Thursday, January 10, 13

  • The argument for existence was really general; rigid foldings exist for these patterns too!

    BGC 2011Thursday, January 10, 13

  • Concentric hexagons:

    BGC 2011Thursday, January 10, 13

  • Concentric octagons:

    BGC 2011Thursday, January 10, 13

  • Concentric decagons:

    BGC 2011Thursday, January 10, 13

  • z = Im(x+ iy)

    nn1

    Rob Kusner suggested

    However, at long distances, z doesnt seem to scale the right way...?

    n d

    2 2

    3 ~1.33

    4 ~1.17

    5 ~1.11

    z rd

    This works when 2n=4, and the branch structure seems to work for all pleated 2n-gons!

    nonalternating BGC 2011Thursday, January 10, 13

  • Other initial conditions....

    There are limiting polygon shapes:

    BGC 2011Thursday, January 10, 13

  • Constant ratio offsets move us in shape space nicer:

    BGC 2011Thursday, January 10, 13

  • Vertex trails of the normalized polygons

    BGC 2011

    A cycle!

    Thursday, January 10, 13

  • Two quick applications

    Thursday, January 10, 13

  • Miura-ori

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Extra creases in Miura-ori: How many degrees of freedom?

    Wei et al, arXiv:1211.6396Thursday, January 10, 13

  • Wu, You, Proc. R. Soc A 2011

    Balkcom et al, 2009

    Rigidly folding a shopping bag?

    Thursday, January 10, 13

  • Wu, You, Proc. R. Soc A 2011

    Balkcom et al, 2009

    Rigidly folding a shopping bag?

    Thursday, January 10, 13

  • Wu, You, Proc. R. Soc A 2011

    Thursday, January 10, 13

  • Ive presented butterflies as a useful tool for understanding the possible rigid motions of folding patterns:

    They should also be useful as atoms for understanding bending energies as well!

    The Future:

    Thursday, January 10, 13

  • Thanks!

    2011 IMA workshop on strain induced shape formation

    Acknowledgements:

    NSF DMR05-47230

    Chris Santangelo, Marcelo DiasRob Kusner

    Thursday, January 10, 13