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2016-05-31

Analyzing the Stability Properties of Kaleidocycles Analyzing the Stability Properties of Kaleidocycles

Clarke A. Safsten Brigham Young University - Provo

Travis B. Fillmore Brigham Young University - Provo

Andrew E. Logan Brigham Young University - Provo

Denise M. Halverson Brigham Young University - Provo

Larry L. Howell Brigham Young University - Provo

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BYU ScholarsArchive Citation BYU ScholarsArchive Citation Safsten, Clarke A.; Fillmore, Travis B.; Logan, Andrew E.; Halverson, Denise M.; and Howell, Larry L., "Analyzing the Stability Properties of Kaleidocycles" (2016). Faculty Publications. 1613. https://scholarsarchive.byu.edu/facpub/1613

This Peer-Reviewed Article is brought to you for free and open access by BYU ScholarsArchive. It has been accepted for inclusion in Faculty Publications by an authorized administrator of BYU ScholarsArchive. For more information, please contact scholarsarchive@byu.edu, ellen_amatangelo@byu.edu.

C. SafstenDepartment of Mathematics,

Brigham Young University,

Provo, UT 84602

T. FillmoreDepartment of Mathematics,

Brigham Young University,

Provo, UT 84602

A. LoganDepartment of Mathematics,

Brigham Young University,

Provo, UT 84602

D. HalversonDepartment of Mathematics,

Brigham Young University,

Provo, UT 84602

L. HowellDepartment of Mechanical Engineering,

Brigham Young University,

Provo, UT 84602

Analyzing the StabilityProperties of KaleidocyclesKaleidocycles are continuously rotating n-jointed linkages. We consider a certain classof six-jointed kaleidocycles which have a spring at each joint. For this class of kaleido-cycles, stored energy varies throughout the rotation process in a nonconstant, cyclicpattern. The purpose of this paper is to model and provide an analysis of the storedenergy of a kaleidocycle throughout its motion. In particular, we will solve analyticallyfor the number of stable equilibrium states for any kaleidocycle in this class.[DOI: 10.1115/1.4032572]

Keywords: kaleidocycle, stable states, energy minima, compliant mechanisms,Bricard 6R

1 Introduction

Kaleidocycles represent a class of mechanisms which havepotential in applications of engineering design due to their uniquecombination of characteristics: they are continuously rotatablemechanisms, they have the ability to possess multiple stable equi-librium states with a variety of characteristics, and they can bedesigned as compliant mechanisms. An understanding of kaleido-cycle behavior, especially the ability to characterize their stableequilibrium properties, has the potential of leading to new com-pact devices not previously possible. This can be particularly use-ful in applications, such as switches, relays, and autonomoussensors. They also show promise as a way of improving durabilityin applications where detents are currently used to achieve multi-ple stable states. Figures 1 and 2 show six-jointed kaleidocycles.

In this paper, we develop analytical solutions for finding thenumber of stable states of six-jointed kaleidocycles with respectto a range of geometric parameters. Associated theorems and theirproofs are presented to support the analysis, and potential engi-neering applications are listed. The work is inspired by prelimi-nary work that numerically approximated the solutions [1].

The popularization of the kaleidocycle is attributed to artistWallace Walker and mathematician Schattschneider whopublished the book M. C. Escher Kaleidocycles in 1977 [2].Kaleidocycles have been an object of fascination ever since, astheir continuous rotation without articulating hinges is accom-plished in a visually complex pattern.

We consider six-jointed kaleidocycles consisting of a ring oftetrahedra for which the tetrahedra are all congruent. We alsorequire that each tetrahedron be joined along one edge to a con-gruent edge of another tetrahedron. Having a single degree-of-freedom, the position of a kaleidocycle through its rotation can bemeasured by the angle between each consecutive pair of tetrahe-dra. The symmetry of the kaleidocycles causes each of the sixangles to have one of just two mutually dependent values. Thesevalues alternate as we move around the kaleidocycle. This

Fig. 1 A compliant-mechanism kaleidocycle that uses polypro-pylene “surrogate folds” to create the function of the rotatingjoints and the torsion springs

Fig. 2 An n 5 6 kaleidocycle

Contributed by the Applied Mechanics Division of ASME for publication in theJOURNAL OF APPLIED MECHANICS. Manuscript received September 2, 2015; finalmanuscript received January 21, 2016; published online February 10, 2016. Assoc.Editor: Shaoxing Qu.

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symmetry enables mechanism mobility and ensures that there isone degree-of-freedom.

To model its kinematics, a kaleidocycle’s tetrahedra can bemodeled as rigid links (the solid lines shown in Fig. 3) and itsjoints as kinematic pairs [3,4], resulting in the mechanism illus-trated in Fig. 4. This is also known as a Bricard 6R mechanism[5–7], and its motion can be modeled using methods employed inthe analysis of spatial mechanisms [8–11].

A kaleidocycle is said to be realizable if and only if it is con-structible and completely rotatable without self-intersections. Weare concerned only with realizable kaleidocycles in this paper.The realizable kaleidocycle is infinitely rotatable.

A structure is said to be monolithic if it can be folded from asingle sheet of material. Kaleidocycles can be monolithic, elimi-nating the need for articulating hinges. Rather, energy is stored inthe springlike flexures which lay on the edges between tetrahedra.Kaleidocycles folded from a single sheet fall into the category ofcompliant mechanisms [12]. The kaleidocycle shown in Fig. 1 is amonolithic compliant mechanism.

As the kaleidocycle rotates, the joints connecting tetrahedrabend, storing and releasing energy. Unlike most compliant mecha-nisms, the stored energy at any position of the kaleidocycle is thesame as when it is in that same position after one more completerotation. Kaleidocycles thus violate a long-held assumption thatmonolithic compliant mechanisms were incapable of continuousrotation.

The stored energy changes throughout the rotation cycle of thekaleidocycle. Because the stored energy is not constant duringrotation, there exist mechanism positions where the energy storedis a local minimum or maximum. The local minima indicate stableequilibria in the rotation, while the local maxima indicate unstable

equilibria. The energy is being stored as strain energy in the bend-ing joints, and they can be modeled as springs with properties thatare functions of the geometry and material properties. The mathe-matical model we use in our analysis assumes springlike hingessuch as those fashioned in compliant mechanisms [13–17]. Thebehavior is similar to that experienced in carton folding [18], onlyhere elastic deflections are assumed. A kaleidocycle’s geometricproperties, along with the properties of the springs, determine thenumber of stable equilibrium states it will exhibit. Many optionsexist for creating bistable mechanisms, but that is not the case forsingle degree-of-freedom multistable systems, which are more dif-ficult to synthesize and analyze. We explore how to analyticallydetermine the number of stable states for a given kaleidocycle.

Multistable mechanisms possess multiple stable equilibriumpositions within their range of motion [19–21]. This characteristiccan be helpful in mechanisms where distinct positions are desired,such as latches [22], surface curvature control [23], relays [24],containers [25], and threshold sensors [26], and no power isrequired to hold a given position, only to transition between posi-tions. The kaleidocycle is not the only paper mechanism to haverecently inspired the study of multistable systems [27–30].

Folded systems, such as the kaleidocycle, origami [31,32], andpop-up mechanisms [33], have inspired systems as diverse as me-chanical metamaterials [34,35], deployable solar panels [36,37],crash boxes [38], deployable masts [39], self-folding machines[40,41], and deployable shelters [42].

2 Preliminaries

2.1 Some Geometry. A tetrahedron can be constructed as fol-

lows: Let S1 ¼ a1b1 and S2 ¼ a2b2 be segments such that the

lines containing them, L1 ¼ a1b1

!and L2 ¼ a2b2

!, are skew. Then,

the tetrahedron s is the join of S1 and S2, so that

s ¼ S1 � S2 ¼[

x2S1 ; y2S2

xy

The direction vectors d1 ¼ a1b1 and d2 ¼ a2b2 differ by anangle a. The value of a is a key property of kaleidocycles. SinceL1 and L2 are skew, there is a nonzero minimal distance betweenL1 and L2. Moreover, there exists a unique segment r joining L1

and L2 that realizes the distance between L1 and L2. The tetrahe-dron s may or may not contain r, but regardless, it is an important

Fig. 4 How h, /, and a are defined from the core segments and hinge lines

Fig. 3 The core segments, with hinge lines indicated

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feature of the kaleidocycle. We refer to r as the core segment ofs. Figure 5 gives a visualization of this construction.

Now that we have constructed a tetrahedron, we can create fiveduplicates of s, three of which are mirror images of the first and con-nect them to form a kaleidocycle as shown in Fig. 2. In particular,note that every tetrahedron is connected to the next along one of theedges. Furthermore, the edges along which adjacent tetrahedra arejoined are congruent edges between tetrahedra. Finally, the tetrahe-dra are connected such that adjacent tetrahedra mirror one another.

If there are any self-intersections in the resulting kaleidocycle, it isnot realizable. Even if there are no self-intersections initially, theremay be intersections later in the rotation process, making the kaleido-cycle not realizable. An example of a kaleidocycle that is realizable isone that satisfies: (1) a ¼ p=2, (2) S1 and S2 are centered on the end-points of the core segment r, and (3) r is longer than both S1 and S2.

2.2 The Bricard 6R. A kaleidocycle is made up of tetrahedrajoined by flexures or articulating hinges. Because all the tetrahe-dra are congruent, the lengths of each of the core segments mustnecessarily be equal. The core segments of a kaleidocycle areillustrated in Fig. 3. All the six core segments together form alinkage that is included in the class of Bricard 6R linkages [6].

A Bricard 6R is an overconstrained linkage of six segments andjoints in a ring which, by virtue of being overconstrained, was thoughtto be immobile according to the Kutzbach Mobility Criterion. Bri-card, however, showed the criterion to be sufficient, though not a nec-essary condition for mobility by providing his 6R linkage as acounterexample. Indeed, the Bricard 6R is mobile and possesses theproperty of infinite rotation [5]. Although Bricard considered a widerange of 6R linkages, a subclass of these linkages corresponds to thekaleidocycles in this paper. Thus, Bricard’s work on 6R linkages laysfoundational ideas for our study of kaleidocycle motion [8].

2.3 Angle Relations. Recall that the angle between vectorslying along the joining edges of tetrahedra is denoted by a. Becauseall the six tetrahedra of a kaleidocycle are congruent, a given kalei-docycle corresponds to a single value a. It should also be noted thatunless a ¼ p=2, the angle a could be measured as its supplemen-tary angle, so that we can choose a > p=2 or a < p=2 on the samekaleidocycle. We will see that the stability properties we examinehere are equivalent for both cases. For a kaleidocycle to be realiz-able, it is a necessary (though not sufficient) condition that

p3< a <

2p3

As stated earlier, the symmetry of the kaleidocycles we considerhere allows us to divide the six joints into two sets of three jointseach. Starting from an arbitrary joint in the core segment structure,every other angle (to a total of three) has a value h, while theremaining three have value /. These angles are the angles betweenadjacent core segments measured on the exterior as shown in Fig.4. The relationship between h, /, and a is given by the relation [7]

g½a�ðh;/Þ ¼ ðcosh� coshcos/� 1þ cos/Þcos2a

þð2sinh sin/Þcosa�ðcoshþ coshcos/þ cos/Þ ¼ 0

(1)

The angles h and / are important in measuring the rotational stateof the kaleidocycle.

Given a, we would like to solve for / in terms of h. Note thatfor almost every realizable h, there are two / values satisfyingEq. (1). Thus, two functions will be required to represent the rela-tion between / and h. We call these functions cþ½a�ðhÞ andc�½a�ðhÞ. In particular,

/ ¼ c6½a�ðhÞ ¼ 6C½a�ðhÞ þ T½a�ðhÞ (2)

where

C a½ � hð Þ ¼ arccos � cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ

!

and

T½a�ðhÞ ¼ Arg½ð�cos2að1� coshÞ þ ð1þ coshÞÞ � ið2 cosa sinhÞ�

We give a full treatment of the derivation of Eq. (2) in theAppendix.

We can use Eq. (2) to reduce functions of the form f ðh;/Þ to afunction of h only, f ðh;/ðhÞÞ. We will abuse notation and simplywrite f ðhÞ when convenient, but we still must take into accountboth forms of Eq. (2). We obtain both plus and minus branches ofc6½a�ðhÞ, which we will denote by

fþðhÞ ¼ f ðh; cþ½a�ðhÞÞ

Fig. 5 A visualization of the geometric construction of the tetrahedron. The lines L1 and L2

are skew. The segments S1 5 a1b1 and S2 5 a2b2 lay on L1 and L2, respectively. The join of S1

and S2 is the tetrahedron s. The dashed segment r is the core segment, whose length is theshortest path between L1 and L2.

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and

f�ðhÞ ¼ f ðh; c�½a�ðhÞÞ

We will see in Theorem 4.5 that

h;/ 2 �arccoscos2aþ 1

2cos2a� 1

0@

1A; arccos

cos2aþ 1

2cos2a� 1

0@

1A

24

35

(3)

It is easily verified that the stability properties of the kaleido-cycle are the same for both a and p� a. In particular,

g½a�ðh;/Þ ¼ g½p� a�ð�h;/Þ ¼ g½p� a�ðh;�/Þ

Thus, the graph of g½a�ðh;/Þ ¼ 0 is a reflection of g½p� a�ðh;/Þ ¼0 across either the line h¼ 0 or / ¼ 0. This is demonstrated in Fig. 6.

2.4 Stored Energy. Each joint in a kaleidocycle acts as ifrestrained by a torsional spring which stores energy according tothe equation

E bð Þ ¼ kb

2b� b0ð Þ2 (4)

where kb is the torsional spring coefficient, b is the angle of thejoint, and b0 is the joint angle at which the spring holds noenergy.

Linear spring behavior is assumed in this work. The spring func-tion may originate from any number of sources, including an actualtorsion spring connected at a pin joint. The most likely origin of theresistance to motion is from the deflection of flexures that deflect toachieve the motion. Relating flexure behavior to spring constants iswell documented in the compliant-mechanism literature [12,13]. Fora traditional kaleidocycle, the spring would be associated with thecrease’s resistance to deflection and may be from folding paperlikematerials [18] or thin sheets of polymers or metals [43]. Less obviousflexures are also possible, such as the polypropylene compliant-mechanism kaleidocycle shown in Fig. 1. In this example, complianttorsion hinges (also known as surrogate folds [44]) enable themotion. Self-intersection is avoided by ensuring that the structure fitsin the tetrahedral volumes described for the traditional kaleidocycle.The linear spring assumption may be invalid if the flexure materialsare strained into the nonlinear range. This can be minimized by usingthin materials at the hinges to reduce the strain. For cases with mod-erate nonlinearities, an approximate linear assumption is still valua-ble for approximating the multistable behavior because the methodrelies on the location of local minimums and maximums of potentialenergy rather than magnitudes. For cases with highly nonlinear mate-rials, spring functions would be used to replace the spring constants.

3 The Case of Two Alternating Sets of Identical Springs

In this section and throughout the next, we assume that thesprings at every other joint in our kaleidocycle are identical. In

Sec. 5, we will see that the results of this case directly apply to thegeneral case in which the springs may differ.

3.1 The Energy Function. Then, the energy of the entirekaleidocycle system is

E h;/ð Þ ¼ kh

2h� h0ð Þ2 þ k/

2/� /0ð Þ2 (5)

where kh is the effective torsional spring constant of the three hjoints, k/ is the effective torsional spring constant of the three /joints, h0 is the angle of the joint at which the h springs holds noenergy, and /0 is the angle of the joint at which the / springsholds no energy.

Note that kh and k/ are the effective spring constants, meaningtheir values account for all the three springs on h joints or / joints,respectively. We consider only positive kh and k/.

We will see that the individual values of kh and k/ are not asimportant as the ratio

r ¼ kh=k/

3.2 Intrinsic and State Variables. The six variables associ-ated with our kaleidocycles are: a, r, h0, /0, h, and /. The varia-bles a, r, h0, and /0 are properties of a kaleidocycle determinedwhen it is constructed and normally are fixed after the construc-tion. These variables are known as intrinsic variables. The inter-dependent variables h and / change as the kaleidocycle rotates.Each one determines the current position or state of the kaleido-cycle. Hence, they are called state variables.

We will explore points at which energy is a local minimum in hand / variables for given values of a, r, h0, and /0. The points atwhich the local minima occur correspond to the stable states ofthe kaleidocycle.

3.3 Kaleidocycle Notation. We denote a kaleidocycle withintrinsic variables a, r, h0, and /0 by

K½a; r�ðh0;/0Þ

We will find this notation useful when we consider collections ofkaleidocycles where a and r are fixed, but where h0 and /0 arevaried.

4 Results

4.1 Motivation. Evans et al. [1] demonstrated numeric solu-tions to the number of stable states of the kaleidocycleK½p=2; 1�ðh0;/0Þ. The result is the star plot in Fig. 7. Each regionin the h0/0-plane corresponds to a collection of kaleidocycleswith an equal number of stable states. The shape of the star wascalculated numerically. The calculations were numerically inten-sive, and each plot required hours of computation time to gener-ate. This is one reason we endeavor to find analytic resultssupporting the numeric results. An analytic method for generating

Fig. 6 The relation g½a�ðh;/Þ5 0

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plots like the plot in Fig. 7 will improve efficiency when perform-ing calculations. We also seek to find a general function for thistype of plot for arbitrary r and a so that we can analyze a broaderclass of kaleidocycles.

Our goal is to derive parameterizations of the curves whichbound the regions in Fig. 7. We denote the star-shaped curve orstar plot for given values of a and r as S½a; r�.

4.2 Lagrange’s Method of Optimization. Because the stablestates of a kaleidocycle occur at energy minima, we will attemptto minimize the energy stored in a kaleidocycle with respect to hand /. We will apply Lagrange’s method of constrained optimiza-tion since h and / are interdependent.

Note that the energy of a kaleidocycle Eðh;/Þ, given in Eq. (5),is a function of two variables subject to the constraint gðh;/Þ ¼ 0,given in Eq. (1). The method of Lagrange indicates that if Eðh;/Þis at an extremum subject to gðh;/Þ ¼ 0, then

rEðh;/Þ � rg½a�ðh;/Þ ¼ 0 (6)

If we simplify the above equation, we get the relation

khðh� h0Þg/ ¼ k/ð/� /0Þgh

We see that we can find h0 as a relation in terms of h, /, and /0.We solve for this relation in terms of two functions N6½/0�ðhÞ

h0 ¼ N6 /0½ � hð Þ ¼ h� c6 a½ � hð Þ � /0

� � 1

r

� �gh

g/(7)

Sometimes we will not specify the branches explicitly and simplywrite

N /0½ � hð Þ ¼ h� c a½ � hð Þ � /0

� � 1

r

� �gh

g/

We can plot N6½/0�ðhÞ for fixed /0, r, and a, as shown inFig. 8. The plot in Fig. 8 shows that for every h 6¼ 0, there existtwo h0 values such that the kaleidocycle is in an equilibrium stateat h. Consider a horizontal line in the hh0-plane, as shown inFig. 8. (The particular line shown in Fig. 8 represents the kaleido-cycle K½p=2; 1�ðp=20; p=20Þ.) The number of times the linecrosses the plots indicates the number of equilibrium positionspossessed by the kaleidocycle K½a; r�ðh0;/0Þ. Note that an

equilibrium position can be either stable or unstable. We need todistinguish the stable solutions from the other solutions.

4.3 Stable and Unstable Equilibrium Values. An exampleof the energy stored in the kaleidocycle as a function of h is givenin Fig. 9. This plot has eight local extrema, four local maxima,and four local minima. We are only interested in the minima, sowe will utilize the second derivative test to determine whichextrema are minima, as mentioned previously. Figure 8 is a graphof the function h0 for which there is an equilibrium at h, but itmakes no distinction between stable and unstable equilibria. Wewill show in Theorem 4.6 that the second derivative of energywith respect to h is, in fact, directly proportional to the first deriva-tive of N with respect to h. By excluding the sections of the plot inFig. 8 with nonpositive slope, we are left with that part of thegraph of h0 which represents a stable equilibrium at h, as shown inFig. 10.

The summary of this section will be two results

(1) Theorem 4.5 identifies the range of h and / values through-out the motion of the kaleidocycle.

(2) Theorem 4.6 determines that d2E=dh2 ¼ khðdN=dhÞ assum-ing that / is given by c6ðhÞ in Eq. (2).

In preparation, we introduce a few more notations.Remark 4.1. Note that g½a�ðh;/Þ is periodic in h and / with a

period of 2p. Hence, any solution to g½a�ðh;/Þ ¼ 0 in R2 is amember of an equivalence class of solutions. The equivalence

Fig. 8 A plot of N6½p=20�ðhÞ for K ½p=2; 1�ðh0; p=20Þ. The kaleido-cycle K ½p=2;1�ðp=20;p=20Þ appears to have eight equilibria. Thetwo plots correspond to the two different branches of N6.

Fig. 9 Energy as a function of h over the full rotation forK ½p=2;1�ðp=30;p=30Þ

Fig. 7 This is the region plot generated numerically for allkaleidocycles K ½p=2; 1�ðh0;/0Þ. The plot shows the number ofstable states based on h0 (x-axis) and /0 (y-axis).

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relation is given by ðh0;/0Þ 2 ½ðh;/Þ� if h0 � h mod 2p and/0 � / mod 2p. Therefore, any solution to g½a�ðh;/Þ in R2 is inan equivalence class of solutions uniquely represented by a solu-tion in ½�p; pÞ � ½�p;pÞ.

DEFINITION 4.2. Let G½a� denote the points in the h/-plane of theconstraint curve g½a�ðh;/Þ ¼ 0 given by

G½a� ¼ fðh;/Þ 2 ½�p;pÞ � ½�p;pÞ : g½a�ðh;/Þ ¼ 0g.DEFINITION 4.3. We define dom½a� to be the closed interval

dom a½ � ¼ �arccoscos2aþ 1

2

cos2a� 1

� �; arccos

cos2aþ 12

cos2a� 1

� �� �

and domðaÞ to be the open interval

dom að Þ ¼ �arccoscos2aþ 1

2

cos2a� 1

� �; arccos

cos2aþ 12

cos2a� 1

� �� �

The following proposition will support the other results in thissection. We have relegated the proof of this proposition to theAppendix, as the argument is a bit tedious.

PROPOSITION 4.4. Let a 2 ðp=3; 2p=3Þ. If ðh;/Þ 2 G½a� andg/½a�ðh;/Þ ¼ 0, then

h ¼ 6arccoscos2aþ 1

2

cos2a� 1

� �

We are now ready to prove the major theorems which allow usto differentiate between stable and unstable equilibria.

THEOREM 4.5. G½a� � dom½a� � dom½a�.Proof. Note that g½a�ðh;/Þ is continuous and has continuous

partial derivatives. By the continuity of g½a�; G½a� is closed in½�p; p� � ½�p;p�, and so it is compact. Let ph : G½a� ! R be theprojection function so that

phðh;/Þ ¼ h

This function is clearly continuous, and it is defined on a compactset, so there must exist a maximizer and a minimizer in G½a�. Letðhmax;/Þ be a maximizing element. If g/½a�ðhmax;/Þ 6¼ 0, then bythe implicit function theorem, there exists a function /ðhÞ definedon an open neighborhood W � R containing hmax such that for allh 2 W; ðh;/ðhÞÞ 2 G½a�. Thus, there exists e > 0, such that

g½a�ðhmax þ e;/ðhmax þ eÞÞ ¼ 0

But then hmax is not a maximizer, and so we have a contradiction.Therefore, if ðhmax;/Þ is a maximizer of ph, then g/½a�ðhmax;/Þ

¼ 0. Similarly, if ðhmin;/Þ is a minimizer, then g/½a�ðhmin;/Þ¼ 0.

Note that g/½a� is defined at every point in G½a�, but byProposition 4.4, it is only zero at two values of h. These two val-ues must correspond to the minimizer and maximizer. In particu-lar, Proposition 4.4 tells us if ðh;/Þ 2 G½a� and g/½a�ðh;/Þ ¼ 0,then

h ¼ 6arccoscos2aþ 1

2cos2a� 1

0@

1A

(8)

Thus,

h 2 �arccoscos2aþ 1

2cos2a� 1

0@

1A; arccos

cos2aþ 1

2cos2a� 1

0@

1A

24

35¼ dom a½ �

By the same argument, interchanging the roles of h and /, thevariable / must also be contained in dom½a�: Therefore,

G½a� � dom½a� � dom½a� �

THEOREM 4.6. Let a 2 ðp=3; 2p=3). Then

d2E

dh2¼ kh

d Ndh

for all h 2 domðaÞ.Proof. Recall from Eq. (11) that

E h;/ð Þ ¼ kh

2h� h0ð Þ2 þ k/

2/� /0ð Þ2

For h 2 domðaÞ, we can express E as a function of h only byletting / ¼ /ðhÞ. Using the notation /h ¼ d/=dh and holding h0

and /0 constant, we see that

d2E

dh2¼ kh þ k//2

h þ k/ /� /0ð Þ/hh

This is the second derivative of energy in h, which is positive forstable equilibria. We obtain from the method of Lagrange multi-pliers from Eq. (6) the function

N /0½ � hð Þ ¼ h� /� /0ð Þ k/

kh

� �gh a½ �g/ a½ �

provided that g/½a� 6¼ 0. But we know by Proposition 4.4 andTheorem 4.5 that

g/½a�ðh;/Þ 6¼ 0

for any h 2 domðaÞ.Observing that g½a�ðh;/Þ ¼ 0 is a level curve of the surface

z ¼ g½a�ðh;/Þ, we invoke the implicit function theorem: At everypoint ðh;/Þ 2 G½a� where g/½a�ðh;/Þ 6¼ 0 (i.e., h 2 domðaÞ), wecan find a local parameterization /ðhÞ such that in a neighborhoodof h

g½a�ðh;/ðhÞÞ ¼ 0

Then,

dg a½ � h;/ hð Þ� �dh

¼ gh a½ � h;/ hð Þ� �

þ g/ a½ � h;/ hð Þ� �

/h a½ � hð Þ ¼ 0

We simplify to get

gh a½ �g/ a½ � ¼ �/h

Fig. 10 N6½/0�ðhÞ for K ½p=2;1�ðh0;p=20Þ. Only the parts of thegraph corresponding to stable equilibria are displayed. The barindicates four stable equilibria.

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Therefore,

d Ndh¼ d

dhhþ /� /0ð Þ k/

kh

� �/h

� �

¼ 1þ k/

kh

� �/2

h þk/

kh

� �/� /0ð Þ/hh

It follows that:

khd Ndh¼ kh þ k//2

h þ k/ /� /0ð Þ/hh ¼d2E

dh2

Hence, the first derivative of N in h is proportional to the secondderivative of E in h by the positive constant kh. �

4.4 Bordered Hessian. At this point, we make a shortdigression to discuss the bordered Hessian matrix. The borderedHessian is a matrix of the second derivatives of a function and firstderivatives of one or more constraints. It is used to determine theconcavity of the function at extremal points with respect to theconstraint(s). As it relates to our problem, the function is energy,Eðh;/Þ and the constraint is, again, g½a�ðh;/Þ. The borderedHessian is given by

H ¼

0@g

@h@g

@/

@g

@h@2E

@h2

@2E

@h@/

@g

@/@2E

@/@h@2E

@/2

0BBBBBBBB@

1CCCCCCCCA

The determinant of this bordered Hessian evaluated at a minimumor maximum gives us information about the stability of oursolutions:

� detðHÞ > 0: unstable solution� detðHÞ ¼ 0: undetermined� detðHÞ < 0: stable solution

One may wonder if an approach performing the bordered Hessianis worth investigating as a method for determining the number ofstable states for a given kaleidocycle. While this is helpful indetermining if a particular solution is stable or unstable, we con-cluded that this approach to the problem of finding the number ofstabilities is more cumbersome than our current approach. How-ever, in problems with higher degrees-of-freedom, this may be amore helpful approach.

4.5 Minima and Maxima of N6½/0�ðhÞ. Figure 10 shows aplot of N6½/0�ðhÞ excluding all the regions of nonpositive slope.Consider a kaleidocycle K½a; r�ðh0;/0Þ constructed with the spe-cific intrinsic value h0, indicated by the horizontal line drawn inFig. 10. The number of intersections between the horizontal lineand the plots indicates the number of stable equilibria for K at thath0 value. If we change the value of h0 under consideration, thenumber of stable states remains the same until we cross a maxi-mum or minimum value of N (see Fig. 11). At this point, the num-ber of intersections, or equivalently the number of stable states,increments or decrements. Likewise, if /0 changes, so do theshapes of the graphs. Consequently, the maxima and minimachange in value, potentially incrementing or decrementing thenumber of stable states for K½a; r�ðh0;/0Þ. Thus, for every valueof /0, there are a collection of local minimum and maximum val-ues of h0 given by N6½/0�ðhÞ at its extrema.

We assemble each pair of ðh0;/0Þ such that N6½/0�ðhÞ attains alocal extreme value of h0. The collection of all such points is thestar plot S½a; r� in Fig. 7 and introduced in Sec. 4.1. The curve

S½a; r� separates the h0/0-plane into regions having a fixed num-ber of stable states.

4.6 Parameterizing the Star Plot. We want to find parame-terizations of the curves comprising the star plot by finding func-tions H06ðhÞ and U06ðhÞ, such that

S½a; r� ¼ fðH06ðhÞ;U06ðhÞÞ j h 2 dom½a�g

Each point ðh0;/0Þ 2 S½a; r� has the property that h0 is an extremevalue of N6½/0�ðhÞ. So we once again apply the method ofLagrange multipliers in order to find extreme values and definethe desired functions. Here, we are merely treating Eq. (7) as afunction of two independent variables subject to the constraintg½a�ðh;/Þ ¼ 0. We want to optimize

N /0½ � h;/ð Þ ¼ h� /� /0ð Þ 1r

gh

g/

recalling that g/ and gh are the partial derivatives of Eq. (1), bothof which have dependence on h, /, and a.

Expanding

0 ¼ rN½/0� � rg½a�

we get

0 ¼ ðN½/0�Þhg/ � ðN½/0�Þ/gh

¼ 1� /� /0ð Þ 1r

ghhg/ � g/hgh

g2/

!g/

� 1

r

gh

g/þ /� /0ð Þ 1

r

gh/g/ � ghg//

g2/

!gh

Hence,

/0 ¼ /�g/ rg2

/ � g2h

� ghhg2

/ � 2gh/ghg/ þ g//g2h

� Substituting in for / using Eq. (2), we define two functionsU06ðhÞ that parameterize the variable /0 for points of S½a; r�

/0 ¼ U06 hð Þ ¼ /6 hð Þ �g/ rg2

/ � g2h

� ghhg2

/ � 2gh/ghg/ þ g//g2h

� (9)

Fig. 11 The bar representing K ½p=2;1�ðh0;p=20Þ now intersectsthe plot at a boundary point

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To parameterize the variable h0 for points of S½a; r�, we simplysubstitute Eq. (9) for /0 in Eq. (7), once again obtaining twofunctions

h0 ¼ H06 hð Þ ¼ h�gh rg2

/ � g2h

� r ghhg2

/ � 2gh/ghg/ þ g//g2h

� (10)

Note that in the equations above, there is a /6ðhÞ dependencein each of the g derivatives not explicitly stated here. Thus, wehave four functions H06ðhÞ and U06ðhÞ which form the coordi-nate pairs for two parametric functions ðH0þðhÞ;U0þðhÞÞ andðH0�ðhÞ;U0�ðhÞÞ. Together, these two parametric functionsdefine the curve S½a; r� in the h0/0-plane. The star plot S½a; r�obtained from these parameterizations in h, assuming r¼ 1 anda ¼ p=2, is shown in Fig. 12.

Figure 13 gives the region plots for variable r and a. The valuesfor each plot are organized in Table 1.

4.7 Interpreting Results. For a given a and r, we can plot itsstar plot S½a; r� that separates the regions where kaleidocycles willhave one, two, three, or four stable states. In particular, if thepoints ðh1

0;/10Þ and ðh2

0;/20Þ lay in the same open connected region

bounded by points in S½a; r�, then kaleidocycles K1½a; r�ðh10;/

10Þ

and K2½a; r�ðh20;/

20Þ have the same number of stable states.

4.7.1 Finding the Number of Stable States. The questionremains, what number of stable states corresponds to each region?It is easy enough to compute the number of stable states for all the18 regions in Fig. 12, but is there a pattern for arbitrary values of rand a? What about the general case? We have observed the fol-lowing pattern that applies in the cases we have investigated: Atextreme values of h0 and /0, a kaleidocycle will always be mono-stable. This is because only one branch of N6½/0�ðhÞ achievesarbitrarily large values of h0. A region is bistable when it shares aboundary with the monostable region. The tristable regions areany regions not already classified as monostable bordering a bista-ble region. Finally, any region bordering only tristable regions arequadstable.

Alternatively, and more precisely, one could simply choose asample point from the region in question, plot the energy curve asin Fig. 9, and count the number of stabilities manually. All the

other points in that region will have the same number ofstabilities.

4.7.2 Varying r and a. We now describe how changes in thevalues of r and a affect the shape of the star plot S½a; r�.

Varying r: Starting at r¼ 1, as r ! 0 or r !1, any quadstableand tristable regions disappear, and it is only possible to obtainmonostable and bistable kaleidocycles. The bistable regionbecomes very large, meaning that most kaleidocycles with smallh0 and /0 will be bistable as r ! 0 or r !1.

Varying a: When the value of a deviates from p=2, the starstretches along the diagonals. The tristable regions get larger asthe quadstable region disappears, but the bistable regions get verylarge as well. So again, almost every kaleidocycle with small h0

and /0 will be bistable as a! p=3 or a! 2p=3.Varying both r and a: Notice that S½p=2; 1� has 4 deg of rota-

tional symmetry and four lines of symmetry: one along each axisand one along each diagonal. If we deviate from ½p=2; 1� in the rvariable, we lose the diagonal symmetries, retaining one rotationalsymmetry. If we deviate from ½p=2; 1� in the a variable, we losethe vertical and horizontal axes of symmetry, but still keep onerotational symmetry. If we deviate from ½p=2; 1� in both the r anda variables, we lose all the axes of symmetry, but still retain asingle rotational symmetry.

4.8 Verification. The multistable behaviors predicted areconsistent with those achieved using the numerical methoddescribed in Ref. [1] and with results achieved in physical testingof hardware described in theses by Evans [45] and Rowberry [46].

5 The Case of Nonidentical Springs

In the general setting of a six-jointed kaleidocycle, we relax theconstraint that the springs at every other joint of the kaleidocycleare identical, so that the spring constants and equilibrium anglesat each joint may differ. The angles at the joints are still governedby the relation

g½a�ðh;/Þ ¼ ðcos h� cos h cos /� 1þ cos /Þ cos2a

þð2 sin h sin /Þcos a� ðcos hþ cos h cos /þ cos /Þ¼ 0

and the range of h and / is the same as before

h;/ 2 �arccoscos2aþ 1

2

cos2a� 1

� �; arccos

cos2aþ 12

cos2a� 1

� �� �

In particular, every other joint will have angle h and the remainingthree joints will have angle /. Then, the energy stored in the jointsof the kaleidocycle is then given by

Eðh;/Þ ¼ Eh1ðh;/Þ þ E/1

ðh;/Þ þ Eh2ðh;/Þ þ E/2

ðh;/Þþ Eh3

ðh;/Þ þ E/3ðh;/Þ

where Ehiðh;/Þ is the energy stored in the ith h-joint, and

E/iðh;/Þ is the energy stored in the ith / -joint.

We must evaluate the critical points and determine which criti-cal points correspond to local minima, or stable states.

If again we assume that our basic model for energy stored in aspring is

Enih;/ð Þ ¼ kni

2n� nið Þ2

where n 2 fh;/g, then

Fig. 12 The parameterized curve S½p=2;1�

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E h;/ð Þ ¼ kh1

2h� h1ð Þ2 þ kh2

2h� h2ð Þ2 þ kh3

2h� h3ð Þ2

þ k/1

2/� /1ð Þ2 þ k/2

2/� /2ð Þ2 þ

k/3

2/� /3ð Þ2

(11)

where khiis the torsional spring constant of the ith h-joint, k/i

isthe torsional spring constant of the ith / -joint, hi is the angle of

the ith h-joint at which the springs holds no energy, and /i is theangle of the ith / -joint at which the springs hold no energy.

An appropriate notation for this kaleidocycle would be

K½a; kh1; k/1

; kh2; k/2

; kh3; k/3�ðh1;/1; h2;/2; h3;/3Þ

5.1 Solving for the Number of Stable States. Solving

rEðh;/Þ � rg½a�ðh;/Þ ¼ 0 (12)

we get the relation

½ðkh1þ kh2

þ kh3Þh� ðkh1

h1 þ kh2h2 þ kh3

h3Þ�g/

¼ ½ðk/1þ k/2

þ k/3Þ/� ðk/1

/1 þ k/2/2 þ k/3

/3Þ�gh (13)

If we define

kh ¼ kh1þ kh2

þ kh3(14)

k/ ¼ k/1þ k/2

þ k/3(15)

Table 1 The values for a and r in Fig. 13

Figures a r

13(a) 5p=12 1/213(b) 5p=12 113(c) 5p=12 213(d) p=2 1/213(e) p=2 113(f) p=2 213(g) 7p=12 1/213(h) 7p=12 113(i) 7p=12 2

Fig. 13 These are the plots of S½a; r � for various values of r and a as described in Table 1. Dif-ferent colored regions represent regions with different numbers of stable states, as follows:Gray—monostable, blue—bistable, yellow—tristable, and red—quadstable.

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h0 ¼kh1

h1 þ kh2h2 þ kh3

h3

kh1þ kh2

þ kh3

(16)

/0 ¼k/1

/1 þ k/2/2 þ k/3

/3

k/1þ k/2

þ k/3

(17)

r ¼ kh

k/(18)

then substituting and solving for h0 we get, as before, Eq. (7)

h0 ¼ N6 /0½ � hð Þ ¼ h� c6 a½ � hð Þ � /0

� � 1

r

� �gh

g/

where

/ ¼ c6½a�ðhÞ ¼ 6C½a�ðhÞ þ T½a�ðhÞ

Because the nonidentical springs can be combined into a singleequivalent spring, the analysis will be the same, including the ver-ification of Theorem 4.6. As a result, the same star graphs apply.In particular, the tetrahedron

K½a; kh1; k/1

; kh2; k/2

; kh3; k/3�ðh1;/1; h2;/2; h3;/3Þ

has the same stable states as K½a; r�ðh0;/0Þ, where r, h0, and /0

are given in Eqs. (14)–(18).Remark 5.1. One implication of this result is that a six-jointed

kaleidocycle, constructed from congruent tetrahedra having differ-ent spring constants and equilibrium angles at each joint, can haveat most four stable states.

5.2 Comparison of Energy Functions. Let E be the energyfunction for

K½a; kh1; k/1

; kh2; k/2

; kh3; k/3�ðh1;/1; h2;/2; h3;/3Þ

given by Eq. (5) and E� be the energy function for

K½a; r�ðh0;/0Þ

given by Eq. (11), where r, h0, and /0 are as in Eqs. (14)–(18).Then,

E h;/ð Þ � E� h;/ð Þ

¼ k1k2 h1 � h2ð Þ2 þ k2k3 h2 � h3ð Þ2 þ k1k3 h1 � h3ð Þ2

2 k1 þ k2 þ k3ð Þ

þ k4k5 /4 � /5ð Þ2 þ k5k6 /5 � /6ð Þ2 þ k4k6 /4 � /6ð Þ2

2 k4 þ k5 þ k6ð Þ

In particular, the energy functions differ by a constant. Thus, thevariations in the energy functions with respect to h and / are iden-tical. Moreover, we observe that E E�, equality holding only ifthe springs are identical. Therefore, the case of identical springsprovides the lowest energy device having a given energy profile.

6 Conclusion

We have shown that it is possible to predict the stability proper-ties of six-jointed kaleidocycles with spring joints. We have pre-sented analytical solutions for finding the stable states and theinfluence of various parameters on system behavior. Theoremsand corresponding proofs support the analysis. The results showpromise for application in a range of engineering systems thatrequire multistable systems, many of which have been identifiedin the paper.

By identifying each kaleidocycle with a point in parameterspace with the corresponding intrinsic variables, we can divideparameter space into regions with similar stability properties.Though this analysis applies to a wide variety of kaleidocycles,it is not exhaustive. We have considered a class of six-jointedkaleidocycles, which represents a subset of the kaleidocycles thatmay exhibit multistable behavior. For example, we could considern-jointed kaleidocycles for n> 6. However, kaleidocycles con-structed with more than six links will have additional degrees-of-freedom which increase the complexity of the kinematics of thesystem. Furthermore, kaleidocycles are a small class of possiblen-jointed linkages, even as a subset of the Bricard 6R family oflinkages. Each n-jointed linkage will have unique relationshipsamong its various rotation angles, similar to the relationshipbetween the variables a, h, and / given by Eq. (1). The analyticalprocess presented in this paper may provide a foundation for stud-ies extending to these larger classes of rotating mechanisms. Theincreased understanding of multistable kaleidocycles can facilitatetheir design for future engineering applications, such as latches,surface curvature control, multistate switches, containers, passivesensors, detent replacements, mechanical valves, and safety devi-ces that require known discrete states.

Acknowledgment

This paper was based on the work supported by the NationalScience Foundation and the Air Force Office of ScientificResearch through NSF Grant No. EFRI-ODISSEI-1240417.

Appendix

A.1 Derivation of Eq. (2).

Here, we show how Eq. (2) can be derived from Eq. (1) via animportant fact

Fact 8.1: For real numbers A, B, and x

A cos xþ B sin x ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

pcos ðx� wÞ

where w ¼ ArgðAþ iBÞ.Equation (1) is the generalized relationship between h, /, and a

g½a�ðh;/Þ ¼ cos2aðcos h� cos h cos /� 1þ cos /Þþ 2 sin h cos a sin /� cos h� cos h cos /� cos / ¼ 0

Let f ¼ cos a and group the terms to match the form in Lemma 8.1

ðf2ð1� cos hÞ � ð1þ cos hÞÞcos /þ ð2f sin hÞsin /

� f2ð1� cos hÞ � cos h ¼ 0

Though it is not immediately obvious, the result is expressedmore simply, in terms of where discontinuities will arise, if wemultiply the above equation by �1, we have

ð�f2ð1� cos hÞ þ ð1þ cos hÞÞcos /� ð2f sin hÞsin /

þ f2ð1� cos hÞ þ cos h ¼ 0

A ¼ �f2ð1� cos hÞ þ ð1þ cos hÞ

B ¼ �ð2f sin hÞ

C ¼ f2ð1� cos hÞ þ cos h

and w ¼ ArgðAþ iBÞ. Then, by Lemma 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

pcosð/� wÞ þ C ¼ 0 (A1)

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We expand and simplifyffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2p

as follows:ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2þB2

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðf2ð1�coshÞ�ð1þcoshÞÞ2þð2fsinhÞ2

q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif4ð1�coshÞ2�2f2ð1�cos2hÞþð1þcoshÞ2þ4f2ð1�cos2hÞ

q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffif4ð1�coshÞ2þ2f2ð1�cos2hÞþð1þcoshÞ2

q

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðf2ð1�coshÞþð1þcoshÞÞ2

q¼ðf2ð1�coshÞþð1þcoshÞÞ

SubstitutingffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2p

and C back into Eq. (A1), we get

ðf2ð1� cos hÞ þ ð1þ cos hÞÞcosð/� wÞþ ðf2ð1� cos hÞ þ cos hÞ ¼ 0

Then,

cos /� wð Þ ¼ � f2 1� cos hð Þ þ cos h

f2 1� cos hð Þ þ 1þ cos hð Þ

Hence,

/ ¼ 6arccos � f2 1� cos hð Þ þ cos h

f2 1� cos hð Þ þ 1þ cos hð Þ

!þ w

where

w ¼ Arg½�ðf2ð1� cos hÞ � ð1þ cos hÞÞ � ið2f sin hÞ�

In other words,

/ ¼ 6arccos � f2 1� cos hð Þ þ cos h

f2 1� cos hð Þ þ 1þ cos hð Þ

!

þ Arg½�f2ð1� cos hÞ þ ð1þ cos hÞ � ið2f sin hÞ�

or

/ ¼ 6arccos � cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ

!

þ Arg½�cos2að1� cos hÞ þ ð1þ cos hÞ � ið2 cos a sin hÞ�

A.2 Helpful Lemmata

In the proof of Proposition 4.4, we make use of the followinglemmata.

LEMMA 8.2. Let a 2 ðp=3; 2p=3Þ. Then, G½a� ð�p;pÞ�ð�p;pÞ.

Proof. By definition, we already know that G½a� ½�p; pÞ�½�p; pÞ. We assume by way of contradiction that there is a pointðh;/Þ 2 G½a� such that h ¼ �p or / ¼ �p. Without loss of gener-ality, suppose that h ¼ �p. Then,

g½a�ð�p;/Þ ¼ 0

implies that

cos2að�1þ cos /þ cos /� 1Þ þ 1þ cos /� cos / ¼ 0

It follows that:

cos2a ¼ 1

2 1� cos /ð Þ

Note that, according to this expression, the minimum possiblevalue of cos2a is 1/4, which corresponds to a ¼ p=3 or a ¼ 2p=3,neither of which is in our domain. Any other solution wouldrequire a < p=3 or a > 2p=3, which is prevented by our hypothe-sis. Thus, h 6¼ �p. Since the roles of h and / are interchangeablein g½a�, we can also conclude that / 6¼ �p. �

LEMMA 8.3. Let a 2 ðp=3; 2p=3Þ. Then, for all ðh;/Þ 2 G½a�

ð�cos2að1� cos hÞ þ ð1þ cos hÞÞ2 þ ð2 cos a sin hÞ2 6¼ 0

Proof. Assume that there exists ðh;/Þ 2 G½a�, such that

ð�cos2að1� cos hÞ þ ð1þ cos hÞÞ2 þ ð2 cos a sin hÞ2 ¼ 0

Then, any solution is a solution to the system of equations

�cos2að1� cos hÞ þ 1þ cos h ¼ 0

2 cos a sin h ¼ 0

The only solutions require h ¼ �p. However, by Lemma 8.2,there are no points ðh;/Þ 2 G½a� where h ¼ �p. It follows that:

ð�cos2að1� cos hÞ þ ð1þ cos hÞÞ2 þ ð2 cos a sin hÞ2 6¼ 0

for all a 2 ðp=3; 2p=3Þ when ðh;/Þ 2 G½a�. �

A.3 Proof of Proposition 4.4

Proof. By implicit differentiation of Eq. (1), we see that ifg/ ¼ 0, then

cos2a sin /ðcos h� 1Þ þ 2 cos a cos / sin hþ sin /ð1þ cos hÞ ¼ 0

We can rewrite the relation above as

sin /ð�cos2að1� cos hÞ þ ð1þ cos hÞÞ þ cos /ð2 cos a sin hÞ ¼ 0

Using the identity A cos /þ B sin / ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2p

cos ½/� ArgðAþ iBÞ� (Fact 8.1), we get

Rða; hÞcos ½/� Argð2 cos a sin hþ ið�cos2að1� cos hÞþ ð1þ cos hÞÞÞ� ¼ 0

where

Rða; hÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið�cos2að1� cos hÞ þ ð1þ cos hÞÞ2 þ ð2 cos a sin hÞ2

q

As shown in Lemma 8.3, Rða; hÞ 6¼ 0, so we divide out Rða; hÞ toget

cos ½/�Argð2cosasinhþ ið�cos2að1�coshÞþð1þcoshÞÞÞ�¼ 0

Thus,

/ ¼ Arg 2 cos a sin hþ i �cos2a 1� cos hð Þ þ 1þ cos hð Þ� �� �

þ 2nþ 1ð Þp2

where n 2 Z. Substituting the above expression into Eq. (2) for /gives

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6arccos � cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ

!

þ Arg½ð�cos2að1� cosðhÞÞ þ ð1þ cos hÞÞ � ið2 cos a sin hÞ�¼ Arg 2 cos a sin hþ i �cos2a 1� cos hð Þ þ 1þ cos hð Þ

� �� �þ 2nþ 1ð Þp

2

We have arguments of different complex numbers on either sideof the equation, so we simplify by writing

z1 ¼ ð�cos2að1� cos hÞ þ ð1þ cos hÞÞ � ið2 cos a sin hÞ

and

z2 ¼ 2 cos a sin hþ ið�cos2að1� cos hÞ þ ð1þ cos hÞÞ

However, we notice that z2 ¼ i � z1, thus Argðz2Þ ¼ ðp=2ÞþArgðz1Þ. The argument terms cancel. Then,

6arccos � cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ

!¼ mp

where m 2 Z. Simplifying, we get

cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ ¼ 61

We have two cases: the þ1 case and the �1 case.Case 1: Suppose

cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ ¼ 1

It follows that:

cos2að1� cos hÞ þ cos h ¼ cos2að1� cos hÞ þ ð1þ cos hÞ

and

0 ¼ 1

Therefore, this case is not possible.Case 2: Suppose

�1 ¼ cos2a 1� cos hð Þ þ cos h

cos2a 1� cos hð Þ þ 1þ cos hð Þ

Solving for h, we get

h ¼ 6arccoscos2aþ 1

2cos2a� 1

0@

1A:

�

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