When actuating a rigid origami mechanism by applying moments at the crease lines, we often confront the bifurcation problem where it is not possible to predict the way the model will fold when it is in a flat state. In this paper, we develop a mathematical model of self-folding and propose the concept of self-foldability of rigid origami when a set of moments, which we call a driving force, are applied. In particular, we desire to design a driving force such that a given crease pattern can uniquely self-fold to a desired mode without getting caught in a bifurcation. We provide necessary conditions for self-foldability that serve as tools to analyze and design self-foldable crease patterns. Using these tools, we analyze the unique self-foldability of several fundamental patterns and demonstrate the usefulness of the proposed model for mechanical design.

Introduction

In mechanical designs based on rigid origami, the folding motion is often achieved by rotational actuators that apply bending moments at each foldline between facets. This self-folding principle is very powerful in different applications; it can be used to design reprogrammable matter , self-folding machines , or to obtain 3D microstructures based on printing patterns . When using self-folding techniques, we always encounter a common question: Is a given configuration possible to be realized by a set of actuators? If so, how? Indeed, the rigid foldability condition [4,5] is a necessary condition for a rigid origami to be actuated as a mechanism; however, this is not at all sufficient. For example, consider a vertex composed of three mountains and three valleys. This can be actuated into two clearly distinct configurations even though they follow the same MV assignment (Fig. 1). Once the vertex starts to pop up or pop down, it cannot flip to the other side without unfolding everything. This bifurcating nature and multistability of origami vertices is an interesting phenomena attracting the attention of scientists . However, from the viewpoint of mechanical design, we often want to avoid such bifurcations. The objective of this paper is to predict the way the model folds from the flat unfolded state and to enable a design of motion via the computation an appropriate set of applied moments. The example of Fig. 1 tells us that we need extra care in order to achieve such a goal. This is not a straightforward problem, as we can observe that even with a set of proper mountain valley folding moments, we may easily encounter bifurcations at the flat state. This flat state singularity is the very reason that the folding of rigid foldable origami such as the Miura-ori or the Resch pattern is difficult at the beginning but is easy when correctly folded for a finite amount.

In this paper, we will pose the mathematical problem of self-folding: To know if there exists a set of moments applied on each hinge, which we later call a driving force, to enable a desired folding motion of a given crease pattern. We give basic theoretical tools for self-folding with which we can analyze and design rigid origami mechanisms based on self-folding.

We also consider a subproblem of self-foldability: Self-foldability with rotational spring driving forces that we define rigorously later in this paper. In this situation, we have a physical implementation of actuators where each edge has a prescribed target fold angle and independently tries to get closer to the target fold angle when actuated, as if torsion springs are attached to creases. We define a mathematical model of such a system and call it a rotational spring driving force. Rotational spring driving forces can model actual methods used in self-folding contexts, e.g., multilayered shape memory materials [1,2] or polymer gel . If a rigid origami is self-foldable with a rotational spring driving force, this can be a very robust system, programmable without active sensing or instantaneous feed-back control.

In this paper, we concentrate on the problem of finding a rotational spring driving force that uniquely self-folds from and to desired rigid origami states. Is this always possible? If not, can we characterize uniquely self-foldable patterns? We believe that the general problem leads to interesting open problems in mathematics and theoretical computer science. As evidence, we will investigate the self-foldability of some interesting examples that can be helpful to grasp the essence of the self-folding problem.

The Self-Folding Problem

Self-foldability is the problem of asking if a rigid folding path from a flat unfolded state to a 3D-folded state can be actuated using a set of driving force (rotational moment) functions without causing a bifurcation at any state. Here, we ignore the inertial effects; in such a system, the springs and dampers dominate, and thus, we can assume that we instantaneously obtain a critical (angular) velocity proportional to the (moment of) force. Self-foldability is an especially important problem at the flat unfolded state where rigid origami constraints degenerate.

Definition 1 (Configuration Space). For a rigid origami with n creases (edges), a configuration is the set of fold angles ρ1,…, ρn of creases 1,…, n of the crease pattern; this can be represented as a point$ρ=[ρ1,…,ρn]T$in n-dimensional parameter space, where n is the number of creases. A rigid folding, i.e., valid configuration, is a configuration which satisfies a set of constraints derived from isometries (Eq.(1)below) and nonintersection of the facets. The set of rigid foldings from the same rigid origami is a subset of the parameter space and is called the configuration space.

Definition 2 (Continuous Folding). A well-behaved continuous rigid folding$ρ(s)$from a rigid folding$ρ(0)$to$ρ(starget)$is an arc length-parameterized piecewise C1 curve in a valid configuration space in n-dimensional parameter space. Because it is piecewise C1, there are at most two tangent vectors$v+(s0):=lims→s0+(dρ(s)/ds)$and$v−(s0):=lims→s0−(dρ(s)/ds)$at a valid configuration.

Consider a set of well-behaved continuous foldings passing through a configuration. The set of tangent vectors of such foldings at the configuration form a region when projected on the unit sphere, which we call valid tangents.

We believe that “well-behaved folding” defined as a piecewise smooth curve captures any process of an actual folding motion that has only a finite number of singular positions. At singular positions, especially at the origin (the flat, unfolded state), the two velocity vectors defined capture the irreversible nature of self-folding and self-unfolding, i.e., self-unfolding is usually easier than self-folding. Figure 2 illustrates the above definitions. Valid tangents may form a nonconnected region on the unit sphere, and the self-foldability problem is deeply related to the geometry of valid tangents.

Definition 3 (Driving Force). A driving force is a continuous vector field in the parameter space:$f(ρ)=[f1(ρ),…,fn(ρ)]T$. A driving force is conservative if it is the negative gradient of some C1 scalar field$U(ρ)$, i.e.,$f(ρ):=−∇U(ρ)$. We call$U(ρ)$the potential energy.

Also, we call a conservative driving force and potential energy a rotational spring driving force and rotational spring potential energy, respectively, if the potential energy is an additively separable function, i.e., the function can be represented as the sum of functions of each fold angle:$U(ρ)=U1(ρ1)+⋯+Un(ρn)$.

In physical sense, a driving force represents a set of moments applied to the hinges. We call it a “force” in the generalized sense, as an energy gradient in some coordinate system. As already mentioned, rotational spring driving forces yield more robust and easy-to-implement systems of self-folding than general potential energy or nonconservative forces that require instantaneous feedback control.

Definition 4 (Constrained Force). We define the constrained forces along a well-behaved continuous rigid folding $ρ(s)$ to be $f+(s):=v+(s)·f(ρ(s))$ and $f−(s):=v−(s)·f(ρ(s))$. We call the former the forward force and the latter the backward force. If a driving force is conservative, then, $f+(s0)=−lims→s0+(∂U/∂ρ)·(dρ/ds)=−lims→s0+(dU(ρ(s))/ds)$.

Definition 5 (Self-Foldable). A well-behaved continuous folding $ρ(s)$ from a rigid folding $ρ(0)$ to $ρ(starget)$ is self-foldable by driving force $f(ρ)$ if at any point $ρ(s)$ for s ∈ [0, starget), the forward force f+(s) is positive and takes a local maximum among the valid tangents at s. Here, we can observe the continuity of vector directions by calculating an intersection of the configuration space with a sphere of radius ϵ around the point when ε → 0. We call a well-behaved continuous folding $ρ(s)$ uniquely self-foldable if $ρ(s)$ is the only well-behaved continuous folding, that is, self-foldable by f.

Basic Kinematics and the Singularity Issue at the Flat State

A rigid folding is valid if and only if it is piecewise isometric and does not self-intersect. If the origami paper forms a disk, the isometry constraints of a rigid origami can be represented by the identity of a rotational matrix product [9,10] as follows: For each interior vertex with foldline direction vectors represented by $ℓi=[ℓix,ℓiy,0]⊤$ for i = 0, 1,…, n − 1 (mod n), their fold angles ρi must satisfy
$R→=I→,where R→:=R→(ℓ0,ρ0)R→(ℓ1,ρ1)⋯R→(ℓn−1,ρn−1)$
(1)

and $R(ℓi,ρi)$ is an orthogonal matrix representing the rotation by angle ρi about an axis along $ℓi$ passing through the origin. As Eq. (1) is a set of polynomial equations of cosines and sines of the fold angles on a compact domain (−π ≤ ρi ≤ π for all i), the valid set of variables is a closed set. We also define the nonintersection condition to allow the paper to touch (but not penetrate) itself. This implies that the configuration space of an origami disk is a closed set.

Consider a configuration where each crease is folded from the original direction $ℓi$ to a 3D direction $Li=[Lix,Liy,Liz]⊤$. The partial derivative of the left-hand side of Eq. (1) is calculated as 
$∂R∂ρi=[Li×]$
(2)
where $[Li×]$ is the matrix representing the cross product operation
$[Li×]:=[0−LizLiyLiz0−Lix−LiyLix0]$
Therefore, the first-order motion satisfies
$∑i=0,…,n−1ρi.Li=0$
(3)

This gives three equations for each interior vertex of a crease pattern. The configuration space is tangent to at least a ein – 3vin-dimensional linear space for an origami model with ein creases and vin interior vertices. However, in the flat unfolded state of $L=ℓ$, the third row of Eq. (3) degenerates, giving us a higher-dimensional tangent space with at least ein − 2vin dimensions. The flat state forms a connecting point of otherwise separated configuration space components.

Degree-4 Flat-Foldable Single Vertex

Consider a degree-4, flat-foldable single vertex with sector angles α,β,π − α,π − β (Fig. 3). Its fold angles ρ0, ρ1, ρ2, ρ3 form a four-dimensional parameter space, and the relationships between these angles give us a valid configuration space that is the union of the following two modes  (see the Appendix for details, Fig. 3)

$t=[tan ρ02, tan ρ12, tan ρ22, tan ρ32]⊤={[t,−pt,t,pt]⊤mode 1[qt,t,−qt,t]⊤mode 2$
(4)
where p and q are the constants given by
$p=p(α,β)=1−tan α2tan β21+tan α2tan β2$
(5)

$q=q(α,β)= tan α2−tan β2 tan α2+tan β2$
(6)

Notice that $0<|p|<1$ and $0<|q|<1$. The folding modes 1 and 2 are each 1-manifolds embedded in four-dimensional parameter space that intersect only at t = 0 (Fig. 4). We, respectively, call them configuration curves 1 and 2. Note that at t = 0, the valid tangents lie within the two-dimensional space defined by Eq. (3).

Theorem 1. For any rigidly foldable, flat-foldable degree-4 vertex with an arbitrary starting and target configurations, there exists a rotational spring driving force that makes the vertex uniquely self-foldable.

Proof. Let $ρ=[ρ0,ρ1,ρ2,ρ3]⊤$ represent the configuration of the model. Assume by symmetry that the target $ρT$ lies on configuration curve 1
$ρT=[τ0,τ1,τ2,τ3]⊤=[τ0,−τ3,τ0,τ3]⊤$

We call the subsets of curve 1 separated by $ρ=0$ manifolds 1+ and 1−, such that curve 1+ includes $ρT$.

We now claim that if there exists a potential energy function $U(ρ)$ with the conditions that

1. (1)

U monotonically decreases along mode 1 toward the target state.

2. (2)

U monotonically decreases along mode 2 toward the flat state.

then, $U(ρ)$ uniquely self-folds from any state to the target state along the shortest path from the initial and targets shapes.

The proof is as follows. Assume that above conditions are satisfied. If we start from a point on curve 1+, then a continuous rigid folding along the shortest path to target position $ρT$ always has positive forward force f+(s) > 0 because of condition 1. Therefore, this path is uniquely self-foldable by U.

Consider that the initial point is on curve 2. Then, a continuous rigid folding along the shortest path from any point $ρ≠0$ on curve 2 to $ρ≠0$ is uniquely self-foldable by U because of condition 2. Similarly, if we start from a point on curve 1−, the shortest path to 0 is uniquely self-foldable by U because of condition 1.

Once we arrive at $ρ=0$ there are four possible paths on which to travel. Here, the one along curve 1+ is uniquely chosen as the tangent vector v+ because

1. (1)

The tangent direction toward 1+ is strictly energy decreasing as f+(s) > 0 because of condition 1.

2. (2)

The tangent direction toward 1− is strictly energy increasing as f+(s) < 0 because of condition 1.

3. (3)

The tangent direction along curve 2 in either direction is energy increasing since f+(s) < 0 once we move away from $ρ=0$.

Therefore, if we start from a point on curve 2 or 1−, the path from the point to 0 and through curve 1+ to $ρT$ is uniquely self-foldable by U.

Here is a rotational spring potential energy that satisfies conditions 1 and 2
$U(ρ)=12∥ρ−ρT∥2$
(7)

$=∑i=0312(ρi−τi)2$
(8)
We first prove that $U(ρ)$ satisfies condition 1. The energy function on manifold 1 is represented as
$U1(ρ0,−ρ3,ρ0,ρ3)=(ρ0−τ0)2+(ρ3−τ3)2$
(9)
This has a global minimum at the target state. We claim that U1 has no other local minimum. To see this, rewrite U1 as a function of single parameter $t=tan(ρo/2)$ in (−, )
$U1(t)=(2arctant−τ0)2+(2arctanpt−τ3)2$
Differentiating U1 by t, we obtain
$dU1(t)dt=21+t2(ρ0(t)−τ0)+2p1+p2t2(ρ3(t)−τ3)$

Since ρ0(t) − τ0 and ρ3(t) − τ3 has the same sign, this takes the value 0 only at the target state.

Next, we prove condition 2. The energy function on manifold 2 is
$U2(ρ0,ρ3,−ρ0,ρ3)=ρ02+τ02+ρ32+τ32$
This has a global minimum only at $ρ=0$ because
$dU2(t)dt=2t1+t2+2pt1+p2t2$

is zero only at t = 0.

Thus, the potential energy given by Eq. (8) uniquely self-folds along a valid rigid folding path from an arbitrary configuration to $ρ=ρT$.

Equation (8) can be simply realized by using rotational springs with rest angles set to target angles; for edge i, we attach a spring with moment fi = k(τi − ρi) (proportional to the angle difference between current and target states). Here, we use the same stiffness k = 2 for all of four creases. Also, this is just an example. We may construct a different set of driving forces with different nonlinear springs such that it also satisfies conditions 1 and 2.

Unique Self-Foldability From the Flat State

As seen in Theorem 1, the driving force must be carefully designed in order for the self folding to pass through $ρ=0$ without getting caught on any bifurcations of the configuration space manifold. We can generalize the necessary conditions for unique self-foldability at the flat, unfolded state.

Lemma 2 (Perpendicular Constraints). A well-behaved continuous rigid folding$ρ(s)$from the unfolded state$ρ(0)=0$with tangent vectorv+ is uniquely self-foldable only if the driving forcefat the unfolded state is perpendicular to every tangent vector in the valid tangents not connected to the projection ofv+ or −v+ on the unit sphere.

Proof. Consider the spherical projection of valid tangents at the unfolded state, which is a closed set because the configuration space is a closed set. These projected tangents are arc-wise connected to the projection of the vector v+ or −v+. Consider a valid tangent vector va that is not connected to v+ or −v+ and assume, for the sake of contradiction, that va is not perpendicular to f(0), i.e., $va·f(0)≠0$. Because the configuration is in a flat state, flipping all mountains and valleys of the valid folding is also a valid folding by symmetry. Therefore, there are two folding paths with tangent ±va, one of which must make a positive dot product with f, i.e., f+ > 0. Since va is in a closed domain, there is a vector vmax that is arc-wise connected to va that locally maximizes $vmax·f(0)$. Since vmax is not v+ or −v+, there is another self-folding motion, which contradicts the uniqueness of our self folding. □

As a consequence of Lemma 2, we are able to get the following important necessary condition.

Lemma 3 (Infinitesimal Dimension Constraints). Consider an origami model at the flat, unfolded state and the tangentvof a desired well-behaved continuous rigid folding. The desired folding is uniquely self-foldable at the flat state only if the dimension m of the solution space of first-order constraints given by Eq.(3)is strictly larger than the number of dimensions n of the linear space spanned by every tangent vectorva not connected tovor −vvia valid tangents.

Proof. Since the configuration space is tangent to the linear space defined by Eq. (3), we have that m ≥ n. Assume that m = n, then the first-order solution space is exactly the linear space composed of the vectors va. Let $v1,…vn$ be linearly independent vectors spanning this space. Then, v can be written as a linear combination of these vectors. However, if the model is uniquely self-foldable in the direction of v, then, the force f must satisfy $vi·f=0$ for i = 1,…, n, which results in v·f = 0. Therefore, it is not possible to design a driving force f that self-folds in the direction of v such that $v·f>0$.

Lemma 4 (All Positive Constraints). A well-behaved rigid folding$ρ(s)$from the unfolded state$ρ(0)=0$with tangent vectorv+, whenv+ and −v+ are in separate components of valid tangents, is uniquely self-foldable only if the force at the unfolded state forms non-negative dot product, f+ ≥ 0, for every tangent vector connected tov+ via valid tangents.

Proof. Assume that there is a vector va arc-wise connected to v+ through valid tangent but forms negative dot product: va·f < 0. Then, its opposite vector −va, which is connected to −v+ though valid tangents, satisfies −va·f > 0. There is a vector vmax that is arc-wise connected to −va that locally maximizes vmax·f(0). Since vmax is not v+, there is another self-folding motion.

The combination of these necessary conditions are useful to the design of driving forces that make the model uniquely self-fold in a desired way. However, this is not sufficient even locally at the flat state; if the subset of valid tangents that is connected to v+ forms a wiggly boundary, it will have another local maximum of f+. In Sec. 6, we use these conditions to prove or disprove unique self-foldability of a rigid origami based on degree-4 vertices.

Connecting Degree-4 Flat-Foldable Vertices

In this section, we consider the family of 1DOF rigidly foldable origami generated by connecting vertices of Theorem 1. An obvious example of this is a Miura-ori, but it is not restricted enough for our purposes. A large variety of over-constrained mechanisms can be generated  by using the linear relationship between opposite fold angles described in tangent of half angle formulas. Such structures also inherit the same nature of the single vertex having separate modes. This makes the structures reconfigurable and reprogrammable into different shapes, but at the same time this makes it difficult for them to self-fold.

Lemma 5. If the interior vertices of a well-behaved continuous rigid folding form an a × b quadrangular grid, then it has exactly a + b linearly independent vectors within the first-order constraints in the flat state.

Proof. For each interior vertex, the first-order constraint at the flat state is given as
$ρ0.ℓ0+ρ1.ℓ1+ρ2.ℓ2+ρ3.ℓ3=0$
Consider that edge 0 is at the top and 1 is at the left. For arbitrarily given ρ0 and ρ1, there is a solution to ρ2 (bottom) and ρ3 (right) satisfying
$ρ2.ℓ2+ρ3.ℓ3=−(ρ0.ℓ0+ρ1.ℓ1)$

because $ℓ2$ and $ℓ3$ are not parallel. Consider the global model of a quadrangular grid. If we arbitrarily choose $ρ˙$ for the b edges on the top and the a edges on the left, then this will sequentially determine $ρ˙$ for connecting edges. This leaves an (a − 1) × (b − 1) grid with top and left edges assigned with $ρ˙$. This process sequentially determines the infinitesimal folding angle of each edge.

As the most simple case, consider connecting two flat-foldable vertices as in Fig. 5. In this case, each vertex can choose its mode from 1 to 2 as in Theorem 1, whose combination yields four possible modes.

Theorem 6. An origami made of two flat-foldable degree-4 vertices is not uniquely self-foldable from a flat state.

Proof. There are four possible tangent vectors along four possible modes at the flat unfolded state. Consider that the left vertex has coefficients of pL, qL and the right has pR, qR. Then, the four infinitesimal modes at flat state can be represented as
$[1−pL1pL−pR1pR11qL−11qL1qR−11qR11qL−11qL−pR1pR1−pL1pL1qR−11qR]$

where each row represents a seven-dimensional infinitesimal mode. The rank of this matrix is 3. Also, the 3 × 7 subset matrix composed of an arbitrary selection of three rows has rank of 3. Now, assume that one of the modes v is uniquely self-foldable. Then, the left-over tuple of modes form linearly independent vectors of a three-dimensional space. Because of Lemma 5, we have a 2 + 1 = three-dimensional tangent space, and by Lemma 3, v cannot be a unique self-folding mode.

This means that if there is a self-folding, then, there is always another valid self-folding mode. Comparing the numbers of folding modes and infinitesimal modes provides good insight toward this type of problem.

As another example, consider a regular square twist in Fig. 6 with free mountain-valley assignment. In this structure, all vertices share the same angle $α=(π/2)$ and β, and thus, the coefficients are

$p=p(α,β)=q(α,β)=1−tan β21+tan β2$
This does not fold like a conventional square twist (which is not rigidly-foldable), but has distinct six modes as illustrated in the figure. We can obtain these six modes by considering the speed assignments for edges around the center square. Each vertex relates the folding speeds of two adjacent edges. The ratio between the tangent of half the folding angles is either p or $(−1/p)$ depending on the mode chosen for each vertex. Because this forms a closed chain around the square, products of the ratios of four vertices must be 1. As p and $(−1/p)$ can be described as $σsign(p)exp (σ log |p|)$ (σ = ±1), the product of four ratios being 1 is equivalent to obtaining a set of valid signs σi(i+1) (i = 1,…, 4 mod 4) that satisfies
$σ12 log |p|+σ23 log |p|+σ34 log |p|+σ41 log |p|=0σ12σ23σ34σ41=1$
These constraints yield following six variations, which are illustrated in Fig. 6
$(++−−),(−++−),(−−++)(+−−+),(+−+−),(−+−+)$

Theorem 7. A regular square twist model is not uniquely self-foldable from a flat state.

Proof. According to Lemma 5, the number of dimensions of the tangent space at the flat state is 2 + 2 = 4. The non-normalized tangent vectors along the six modes can be represented as
$[1pp2−pp−p1−p2p−p−p21pp2−p11−p2p−p−p21p−pp2−p1pp−p−p21p−p1−p2−p1pp2−p21p−p1−p2p−p1p−1−pp−p1−1−pp−11p−1−p11−1−pp−11p−p]$

whose rank is 4. Removing row 1 or 5 does not change the rank of the matrix. Because of symmetry, any submatrix composed of five arbitrary rows is also rank 4. According to Lemma 3, this cannot uniquely self-fold from the flat state.

On the other hand, if the structure has less symmetry, it may uniquely self-fold. Consider a similar twist fold with vertex ratio p and $(−1/p)$ but with one corner having ratio of −p3 and $(1/p3)$. This will have closure constraints of
$σ12 log |p|+σ23 log |p|+σ34 log |p|−3σ41 log |p|=0σ12σ23σ34σ41=1$

This will give only two solutions (σ12, σ23, σ34, and σ41) = (+++−) or (− − −+), visualized in Fig. 7.

Theorem 8. The irregular square twist in Fig. 7 is uniquely self-foldable from and to arbitrary states by a rotational spring force.

Proof. The configuration space consists of two modes 1 and 2. Consider a target state $ρT=(τ1,…,τ12)T$ in mode 1. Notice that in these two global modes, every vertex also folds in different modes. Here, we construct a potential energy by the summation of potential energy for each vertex, i.e., Eq. (8)
$U(ρ)=∑i=14(ρi−τi)2+12∑i=512(ρi−τi)2$
(10)

In mode 2, the energy function is the summation of functions monotonically decreasing toward the flat-state, which is again monotonically decreasing toward the flat-state. In mode 1, the energy function is the summation of functions monotonically decreasing toward the target state, which is again monotonically decreasing toward the target state. Therefore, the rotational spring energy in Eq. (10) uniquely self-folds to $ρT$ from any configuration.

Equation (10) can be realized by using rotational springs with rest angles set to target angles: for edge i, we attach a spring with moment fi = ki(τi − ρi) (proportional to the angle difference between current and target states). Here, the stiffness for the creases shared by two interior vertices is ki = 4 (i = 1…4), and the stiffness for the creases shared only by one vertex is ki = 2 (i = 5…8).

One may double-check the perpendicularity at the flat state. Non-normalized tangent vectors along two modes at the flat state can be represented as
$v1=[1pp2p3−p3−p1−p2p−p3p21]Tv2=[p3−p2p−1−1−p2−p3pp2−1−p−p3]T$
The driving forces toward first and second mode can be represented as
$f1=[2τ12τ22τ32τ4−τ4−τ2τ1−τ3τ2−τ4τ3τ1]Tf2=[2τ12τ22τ32τ4τ4τ2−τ1τ3−τ2τ4−τ3−τ1]T$
Then, the forces and tangents are perpendicular to each other:
$f1·v2=f2·v1=0$

Degree-6 Vertex

Finally, we come back to the degree-6 vertex shown in Fig. 1. As already mentioned, the driving force toward the correct mountain and valley assignment, $f=(f,−g,f,−g,f,−g)T$ (f, g > 0), at the flat state can self-fold into either of pop-up or pop-down states since both of them share the same mountain-valley assignment, and thus, this driving force will not uniquely self-fold. We will show an insight toward designing the same vertex to uniquely self-fold into one of the states in Fig. 1.

This vertex with six creases has three degrees-of-freedom in a generic state. For simplicity, we assume threefold symmetry and treat it as a 1DOF mechanism. As this is a more restrictive configuration space, unique self-folding in this set-up does not imply actual self-folding; we will discuss the generalization later.

Configuration With Threefold Symmetry.

Assume that the configuration can be represented as $ρ=[ρA,ρB,ρA,ρB,ρA,ρB]T$. Then, the closure constraint (1) can be transformed into the following form:
${Rx(ρA)Rz(60 deg)Rx(ρB)Rz(60 deg)}3=I3$
(11)
where Rx(ρ) is rotation about x axis by ρ and Rz(θ) is rotation about z axis by θ, which can be written as
$Rx(ρ)=[1000 cos ρ−sin ρ0 sin ρ cos ρ]Rz(θ)=[ cos θ−sin θ0 sin θ cos θ0001]$
Equation (11) is equivalent to saying that $R1/3:=Rx(ρA)Rz(60 deg)Rx(ρB)Rz(60 deg)$ is a $ϕ=120 deg$ rotation about some axis.1 According to Rodrigues' Formula, the rotation angle $ϕ$ and the trace of the rotation matrix has the following relation:
$Trace(R1/3)=2 cos ϕ+1=0which is calculated as Trace(R1/3)=54cos ρA cos ρB−sin ρA sin ρB−34(cos ρA+cos ρB)+14$
Then, we obtain
$(3 cos ρA+ρB2−cos ρA−ρB2+2)×(3 cos ρA+ρB2−cos ρA−ρB2−2)=0$
(12)
In order for the configuration to have no self-intersections, the measure of the solid angle formed by the vertex calculated as 2π − 3(ρA + ρB) must exist in [0, 4π]. Therefore
$12≤cos ρA+ρB2≤1$
Equation (12) is satisfied if and only if
$3 cos ρA+ρB2−cos ρA−ρB2−2=0$
(13)

This forms a configuration space composed of two curves intersecting at the flat state (Fig. 8). The folding path can be simplified as follows:

$[ tan ρA4 tan ρB4]={[(2+3)t−t]mode 1[−t(2+3)t]mode 2$
(14)
This folding motion has the following velocity in the flat, unfolded state
$v1|ρ=0=[2+3−1]$
(15)

$v2|ρ=0=[−12+3]$
(16)
This tells us that in order to uniquely self-fold to the pop-down state in mode 1 from the flat state, we must choose a force that is perpendicular to $v2|ρ=0$ by Lemma 2. This means that we should use a force parallel to
$[2+31]$

at the flat state. Thus, a set of weak valley and stronger (approximately 3.73 times stronger) valley force assignments is necessary, instead of a native assignment following that of the target shape (alternating mountains and valleys). In fact, in the symmetric case a proper driving force uniquely self-folds the vertex.

Theorem 9. The regular degree-6 vertex with threefold symmetry constraints is uniquely self-foldable from and to arbitrary states by a rotational spring force.

Proof. Consider that the target state $ρT=[τAτB]T$ is in mode 1, and exists in τA ≥ 0. We may reparameterize:
$[ tan τA4 tan τB4]=[(2+3)tT−tT]$

Thus, if we use the tangent of quarter angle as the axes for the parameter space, the configuration space is represented by straight lines (Fig. 9). As the derivative of $tan(ρ/4)$ with respect to ρ is positive in −180 deg < ρ < 180 deg, the signs of the forward force along a folding path are preserved in this remapped parameter space.

Therefore, our objective is to construct a potential energy that strictly minimizes toward $ρT$ along mode 1 and strictly minimizes to $ρ=0$. We may construct such a rotational spring energy by considering an off-configuration target $ρT′=(τA′,τB′)$, which is, in the remapped parameter space, the intersection point of the ray passing through origin toward the direction of $[2+3,1]T$ and the line passing through $ρT$ perpendicular to the tangent v1 at $ρT$. Such a point always exists because the folding path exists on a straight line.

We construct the spring potential energy which minimizes at $ρT′$
$U=12(tan ρA4−tan τA′4)2+12(tan ρB4−tan τB′4)2$
(17)

This energy is quadratic in this remapped parameter space, and strictly minimizes to $ρ=0$ along mode 2 and strictly minimizes to $ρT$ along mode 1, and thus, energy (17) uniquely self-folds under a threefold symmetry constraint.

Validity Without Symmetry Constraints.

We conjecture that this symmetric path is actually a valid self-folding path even without the symmetry constraint. This can be verified by considering the orthogonal projection fproj of the driving force f to the solution space of first-order constraints. Figure 10 illustrates the crease lines in the folded position. Applying the spherical laws of sines, we obtain that the vector position of the creases Li (i = 0,…,5) in mode 1 can be represented as

$Li={[ cos −ρB2cos π3i, cos −ρB2, sin −ρB2]⊤i:even[ cos −ρA2, cos −ρA2, sin −ρB2]⊤i:odd$
The first-order constraint is given using the 3 × 6 matrix C = [L0,…, L5] as
$Cρ˙=0$
The orthogonal projection matrix to the first-order solution space is given by
$P=[I6−CT(CCT)−1C]$
We can now observe from a numerical plot that
$P[f1Af1Bf1Af1Bf1Af1B]=c[vAvBvAvBvAvB]$

for some c > 0, i.e., the projected driving force is parallel to the symmetric folding mode 1 for any 0 ≤ t ≤ 0.5 and 0 ≤ tT ≤ 0.5. This means that the constrained force v·f1 is maximized in the direction of v = v1, so there is a symmetric self-folding motion along mode 1. Although we did not check if constrained force along v1 is the only local maximum, we believe this is so. If it is, the rotational spring force can uniquely self-fold a degree-6 vertex.

Conclusion

In this paper, we proposed a mathematical model of self-folding, specifically for unique-self folding by rotational spring driving forces. We provided necessary conditions for self-foldability that serve as tools to analyze and design self-foldable crease patterns. Using these tools, we demonstrated several results: a degree-4 flat-foldable vertex is uniquely self-foldable; two-vertex and regular twist models based on these vertices are not uniquely self-foldable, but we can design a nonregular twist, that is, uniquely self-foldable. Also, we demonstrated the self-foldability of a degree-6 vertex using the driving force with alternating strong and weak valleys. We believe that these tools can be a basis for future design methods of mechanisms and robotics based on origami. In particular, each of the examples with self-foldability forms a configuration space branching out to different 1DOF mechanisms, while we can assign driving forces that correctly make it choose one of the target modes. This controllability leads to the design of reprogrammable origami systems that can fold into different mechanisms. However, the characterization of self-foldability is still an open problem, and we would like to explore further in this direction.

Acknowledgment

The first author was supported by the JST PRESTO program and JSPS KAKENHI 16H06106. The second author was supported by NSF grant EFRI ODISSEI 1240441.

Nomenclature

Parameters in Crease Patterns and Folded States
• ein =

number of creases in a crease pattern

•
• $ℓi$ =

normalized direction vector of edge i from the vertex in a flat unfolded state

•
• Li =

normalized direction vector of edge i from the vertex in a folded state

•
• p, q =

coefficients of a degree-4 flat-foldable origami vertex

•
• ρi =

fold angle, i.e., the supplementary angle of dihedral angle at the fold line i

•
• vin =

number of interior vertices in a crease pattern

•
• τi =

target fold angle of fold line i

Variables in the Parameter Space

Variables in the Parameter Space

• $f(ρ)$ =

driving force (a set of applied moments). Vector

•
• f(s) =

constrained force along v, defined by $f(ρ(s))·v(s)$

•
• s =

an arclength parameter of a folding path

•
• t =

a parameter of a folding path

•
• t =

a vector representing the current configuration by tangents of half fold angles. i-th element is $tan 12ρi$

•
• $U(ρ)$ =

potential energy of the applied moments. Scalar.

•
• v(s) =

normalized tangent vector of a folding path in parameter space

•
• $ρ$ =

a vector representing the current configuration. i-th element is ρi

•
• $ρ(s)$ =

a folding path

•
• $ρT$ =

configuration of the target position. i-th element is τi, the target folding angle of fold line i

Appendix: Proof of Eq. (4)

Theorem 10. The configuration space of a degree-4 flat-foldable vertex is the union of Eq.(4).

Proof. For necessity, consider that there exists a folded state, and consider a spherical intersection of the vertex with a unit sphere. This is a four-bar spherical linkage with edge lengths of sector angles.

Now, we can check the symmetry of Eq. (4) by replacing α by π − α or α by β. This means that we can safely assume that α is (one of) the smallest angle(s) of four sector angles.

Also, because the configuration space is symmetric with respect to $ρ=0$, assume that this vertex is popped down as the pop-up state can always expressed by negating all fold angles of pop-down state. From single-vertex rigid foldability , the vertex must have three valley creases ca, cb, cc forming sector angles strictly less than π, and 1 mountain crease cd.

Now, the pop-down state is possible if and only if

1. (1)

α < π − β and the assignment of c0, c1, c2, c3 is V, M, V, V.

2. (2)

α = π − β and the assignment of c0, c1, c2, c3 is V, 0, V, 0.

3. (3)

α < β and the assignment of c0, c1, c2, c3 is M, V, V, V.

4. (4)

α = β and the assignment of c0, c1, c2, c3 is 0, V, 0, V.

where M, V, 0 refer to mountain, valley, and uncreased, respectively. The folding of case 2 satisfies mode 1 with pa = 0. Similarly, the folding of case 4 satisfies mode 2 with pb = 0.

In case 3, the vertex is expressed as a quadrangle on the sphere whose interior is the front side of the vertex. The quadrangle is convex at c1, c2, c3, and concave at crease c0, so the geodesic segment between c0 and c2 lies inside this quadrangle (Fig. 11). Therefore, segment c0c2 divides the quadrangle into two triangle composed of convex angles. Let γ < π denote the length of c0c2. Now, we consider the opposite point $c′3$ of c3. Then, the angle $∠c0c′3c2$ equals π − ρ3 because $c3c′3=π$. Now, triangle c0c1c2 and $c0c′3c2$ are congruent to each other both of them share segment lengths α, β, γ < π. Because these triangle share the same base angles, $ϕB=π−ϕD$ and $ϕC=π−ϕA$. Then, we get that $ρ0=ϕB+ϕC−π=π−ϕA−ϕD=ρ2$.

Now, use Napier's Analogies of spherical trigonometry to get the following relations:
$tan ρ22 tan ρ12= tan ϕB−ϕA2cotπ−ρ12=− sin α−β2 sin α+β2=− tan α2−tan β2 tan α2+tan β2$

Such a folding can be represented by the mode 2 of Eq. (4). Similarly, the folding of case 1 satisfies mode 1 of Eq. (4).

For sufficiency, Eq. (4) satisfies Eq. (1). Therefore, Eq. (4) represents the entire configuration space of the vertex.

1

ϕ = 0 yields solution of flat-folding ρA = ±π and ρB = ∓π, but does not lead to a valid folding because there is no layer ordering.

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