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 Tachi’s 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 Lang’s 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
Mårten 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
Mårten 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,j�1) = pi,j�1
d(Vi,j , Vi,j) = pi,j
d(Vi,j , Vi,j+1) = pi,j�1
BGC 2011Thursday, January 10, 13
nonalternating:
alternating:�n+2 ��n
✏=�a2 + b2p
2k2� a+ 2b
2k3+O(k�4)
kn+2 � k✏
=
p2
k� ✏
k3+O(k�4)
an+2 � a✏
=✏
k3+
a(b2 � a2)p2k4
+O(k�5)
bn+2 � b✏
=b(a2 � b2)p
2k4+
✏(2a2 + ab+ 2b2)
2k5+O(k�6)
�n+2 ��n✏
=�p2(a2 � b2) + ✏
2k2� 3✏(a+ b)
4k3+O(k�4)
kn+2 � k✏
=
p2
k� ✏
k3+O(k�4)
an+2 � a✏
=✏
2k3+
p2a(b2 � a2) + ✏(b� 2a)
2k4+O(k�5)
bn+2 � b✏
=� ✏2k3
+
p2b(b2 � a2) + ✏(a� 2b)
2k4+O(k�5)
Non-alternating Alternating
(I dropped most subscripts when they are n) BGC 2011Thursday, January 10, 13
kn+2 � k✏
⇡p2
kkn ⇡
q2p2✏n+ C
Nonalternating:
�n+2 ��n✏
⇡�p2(a2 � b2) + ✏
2k2
=O(�1/n)
an+2 � a✏
=O(n�3/2)
bn+2 � b✏
=O(n�3/2)
Non-alternating Alternating
BGC 2011Thursday, January 10, 13
kn+2 � k✏
⇡p2
kkn ⇡
q2p2✏n+ C
Nonalternating:
�n+2 ��n✏
⇡�p2(a2 � b2) + ✏
2k2
=O(�1/n)
an+2 � a✏
=O(n�3/2)
bn+2 � b✏
=O(n�3/2)
Non-alternating Alternating
�n = �O(log n)
BGC 2011Thursday, January 10, 13
kn+2 � k✏
⇡p2
kkn ⇡
q2p2✏n+ C
Nonalternating:
�n+2 ��n✏
⇡�p2(a2 � b2) + ✏
2k2
=O(�1/n)
an+2 � a✏
=O(n�3/2)
bn+2 � b✏
=O(n�3/2)
Non-alternating Alternating
�n = �O(log n)
Folding the nonalternating pattern doesn’t approximate a hyperbolic paraboloid! BGC 2011
Thursday, January 10, 13
Non-alternating Alternating
kn+2 � k✏
⇡p2
kkn ⇡
q2p2✏n+ C
Alternating:We can show that an � bn = O(1/k)
�n+2 ��n✏
=�a2 + b2p
2k2� a+ 2b
2k3+O(k�4)
=O(k�3)
=O(n�3/2)
an+2 � a✏
=O(k�3) = O(n�3/2)
bn+2 � b✏
=b(a2 � b2)p
2k4+O(k�5)
=O(k�5) = O(n�5/2)BGC 2011
Thursday, January 10, 13
Non-alternating Alternating
kn+2 � k✏
⇡p2
kkn ⇡
q2p2✏n+ C
Alternating:We can show that an � bn = O(1/k)
�n+2 ��n✏
=�a2 + b2p
2k2� a+ 2b
2k3+O(k�4)
=O(k�3)
=O(n�3/2)
an+2 � a✏
=O(k�3) = O(n�3/2)
bn+2 � b✏
=b(a2 � b2)p
2k4+O(k�5)
=O(k�5) = O(n�5/2)
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
θ=0.1
θ=π/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)
nn�1
⇤Rob Kusner suggested
However, at long distances, z doesn’t 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
I’ve 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