A novel criterion for singularity analysis of parallel mechanisms

Mechanism and Machine Theory 137 (2019) 459–475

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Mechanism and Machine Theory journal homepage: www.elsevier.com/locate/mechmachtheory

Research paper

A novel criterion for singularity analysis of parallel mechanisms Michael Slavutin∗, Offer Shai†, Avshalom Sheffer, Yoram Reich School of Mechanical Engineering, Tel Aviv University, Ramat Aviv 69978, Israel

a r t i c l e

i n f o

Article history: Received 19 December 2018 Revised 2 March 2019 Accepted 2 March 2019

Keywords: Instantaneous screw axis (ISA) Minimal parallel robots Screw theory Singular characterization 3D Kennedy theorem 3/6 and 6/6 Stewart Platform

a b s t r a c t A novel criterion for singularity analysis of parallel robots is presented. It relies on screw theory, the 3-dimensional Kennedy theorem, and the singular properties of minimal parallel robots. A parallel robot is minimal if in any generic configuration, activating any leg/limb causes a motion in all its joints and links. For any link of the robot, a pair of legs is removed. In the resulting 2 degrees-of-freedom mechanism, all possible instantaneous screw axes belong to a cylindroid. A center axis of this cylindroid is determined. This algorithm is performed for three different pairs of legs. The position is singular, if the instantaneous screw axis of the chosen link crosses and is perpendicular to three center axes of the cylindroids. This criterion is applied to a 6/6 Stewart Platform and validated on a 3/6 Stewart Platform using results known in the literature. It is also applied to two-platform minimal parallel robots and verified through the Jacobian; hence demonstrating its general applicability to minimal robots. Since any parallel robot is decomposable into minimal robots, the criterion applies to all constrained parallel mechanisms. © 2019 Elsevier Ltd. All rights reserved.

1. Introduction This paper introduces a novel criterion for singularity analysis of general parallel mechanism. We distinguish between characterization and criterion of a singular configuration. The former provides a geometric description of a parallel robot in a singular position, while the latter checks whether the geometry of a given robot indicates that it is in a singular configuration. Singularity analysis is one of the fundamental issues of parallel mechanisms. Singular configuration is the position where the mechanism gains or loses degrees of freedom. It is known to be one of the main concerns in the analysis and design of mechanisms [1]. Although singularity of parallel mechanisms was studied early in the 1980s, this problem is still very difficult. Besides studying singularity positions, it is also critical to avoid them [2], to escape them as fast as possible [3], and in general, to characterize the geometry of the singularity spaces [4]. Among the methods used in determining singularity positions are the analysis of the Jacobian matrix, obtained from analytical expressions [5,6], or from geometric analysis that uses Grassmann-Cayley algebra [7–10], or using quaternion algebra [11]; motion/force transmissibility analysis [12]; and rigidity matrix analysis [13]. The latter is extensively used in the mathematics community [14] in rigidity theory. Some papers dealt with criteria for singularity of different types of

∗

†

Corresponding author. E-mail addresses: [email protected] (M. Slavutin), [email protected] (A. Sheffer), [email protected] (Y. Reich). Deceased

https://doi.org/10.1016/j.mechmachtheory.2019.03.001 0094-114X/© 2019 Elsevier Ltd. All rights reserved.

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Nomenclature AG DOF ISA PR SP ◦ ∨ ∧ 2D 3D 3/6 SP 6/6 SP S, s S0 , s0

Assur graph Degree(s) of freedom Instantaneous screw axes Parallel robot Stewart platform Reciprocal product Join Meet Two-dimensional Three-dimensional 3/6 Stewart platform 6/6 Stewart platform Direction of a screw or a line Moment of a screw or a line

parallel mechanisms [15], especially based on screw theory. In the works that use Geometric Algebra, first by Tanev [16], the singularity is determined as vanishing of the six-blade, which is yet another form of the Jacobian [17,18]. Although the above methods allow the determination of all types of singularities, they usually do not give the geometrical characterization of singularity positions. Contrary to those methods that analyze the algebraic relations, this article continues the geometrical approach presented in Ref. [19]. This approach looks for a visual description of the singularity position that can subsequently transform into an algebraic form, using any existing approach, be it Grassmann-Cayley algebra, Conformal Geometric Algebra, or vector algebra. Once the description of the singularity position is transformed into algebraic form, it can be used in various numerical analyses, for example, the analysis of the workspace. The geometric analysis, long known for planar mechanisms [20,21], is available for 3 dimensional (3D) mechanisms in special singularity cases only [7]. In this article, the visual geometrical criterion will be transformed into algebraic notation consistently using Plücker coordinates and Grassmann-Cayley algebra as an example. Thus, the geometric criterion developed here can be implemented and tested easily. Since it uses a series of geometrical building steps, its implementation does not require calculating determinants or inverting matrices. The singularity criterion is validated in this paper on the Stewart Platform (SP), which is the most typical parallel mechanism with six degrees of freedom and a known minimal parallel robot [22,23], and on general multi-platform parallel robots. Contrary to the previous paper [19], where the characterization assumes that the 6/6 SP has two lines that cross 4 of the legs of the platform; here, we relax this assumption, and the platform can be in any general configuration. This is due to the property that all the instantaneous screw axes of a 2 DOF (Degrees of Freedom) mechanism belong to a cylindroid, and an algorithm for determining the center axis of this cylindroid of the ISAs of the general four-leg platform, that is introduced in this article. The criterion for the 6/6 SP will be validated on the 3/6 SP for which other geometric criteria are reported in the literature (Section 5). Downing et al. [13], followed by Ben Horin and Shoham [10], proved, relying on Grassmann-Cayley algebra, that all the singularity configurations of the 3/6 SP are delineated to be four planes intersecting at a single point. They showed that all the singular configurations found earlier by Merlet [9] are special cases of this condition. We show that our criterion covers this result. While the criterion is developed and illustrated on minimal parallel robots, it applies to all constrained parallel robots, since any parallel robot can be decomposed into minimal parallel robots [24]. Consequently, this paper presents the first general geometric criterion for singular positions of general constrained parallel robots. This paper is organized as follows. Section 2 provides a brief explanation of the 3D Kennedy theorem. Section 3 introduces the main properties of minimal parallel robots and their relations to Assur Graphs. The proposed novel singularity criterion for parallel robots is introduced in Section 4 by using the reciprocal product (screw theory) and the dual 3D Kennedy theorem, and implemented on the 6/6 SP in the most general configuration. Application of this criterion to planar mechanisms is demonstrated in Section 4.3. Section 5 introduces a comparison to the singularity result that appears in the literature in the context of 3/6 SP singularity. Section 6 demonstrates the application of the criterion to multi-platform robots: one for which no existing result appears in the literature and the other presented by Song et al. [25]. Section 7 concludes the paper. Note that all the figures in this paper, including the geometric entities, were drawn or constructed precisely with a 3D modeling tool and accurately reflect reality. 2. The 3D Kennedy theorem The Aronhold-Kennedy theorem in 3D can be formulated and proved through screw theory [19]. For the sake of completeness, we reproduce here the lemmas and the theorem that will be used later in Section 4. Lemma 1. The reciprocal product of two lines, L1 and L2 , which cross each other is zero [26].

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Lemma 2. Let $ be a proper screw [27] (h = 0, h = ∞) and L a line. Then, any one of the following conditions could be proved by using the other two: a. The reciprocal product of a screw and line L is zero. b. The line of screw $ and line L cross each other. c. The directions of screw $ and line L are perpendicular. With these two lemmas it is possible to prove the 3D Kennedy theorem [19]. Kennedy theorem in 3D [28,29]: for any two screws, $1 and $2 , the relative screw $1, 2 crosses and is perpendicular to the common normal l between the two screws $1 and $2 . The Aronhold-Kennedy theorem in 3D applies to both kinematics and statics. In this paper it is used in statics, while in most of the work reported in the literature, it is used in kinematics [29]. 3. The main properties of minimal parallel robots/Assur graphs This section introduces the concept of minimal parallel robots and their main properties. Definition of a minimal parallel robot: a parallel robot is minimal, if and only if when in any generic configuration, activating any leg/limb causes a motion in all the joints and links of the PR. As seen in Fig. 1a, movement of any inner joint, A1 − A5 , causes a motion in all the joints and links. For example, movement of joint A5 by extending actuator 8 results in a movement of joints A1 − A4 . In contrast, moving joint A4 in Fig. 1b does not cause a motion in the inner joint of the spatial Triad that consists of joints A1 − A3 and links 1–9, so the mechanism in Fig. 1b is not a minimal parallel robot. Since the definition of minimal parallel robots is exactly the definition of Assur Graphs [30], it follows that minimal parallel robots are actually Assur Graphs (AG). The concept of AG was first published as Assur Groups in [31,32] and at that time was widely employed in Russia and other eastern European countries. Assur Graphs are the mathematical reformulation of Assur Groups [30] in terms of rigidity theory and graph theory, known to be the building blocks of any mechanism in 2D and also in 3D. For example, the mechanisms 3/6 SP and 6/6 SP are the known Assur Graphs: 3D bar-joint Triad and 3D body-bar atom, respectively. AGs were proved to possess special properties [30,33]; they have special singularity properties that only they possess, such as: the system is both mobile (has an infinitesimal motion) in all the joints with 1 degree of freedom (1DOF) and has a self-stress, i.e., inner forces in all its elements. In addition, when an AG is in a singular position, removing any link does not affect the motion of all the other links, i.e., its instantaneous screw axis (ISA) remains the same. This property is widely used in the singular criterion proposed in the paper for minimal parallel robots. Any general non-minimal constrained parallel robot is a composition of Assur Graphs. For example, the robot in Fig. 1b is a composition of a spatial Dyad that consists of joint A4 and links 10–12, and of a spatial Triad that consists of joints A1 − A3 and links 1–9. All the singularities of the general robot are described by the singularities of its minimal components. This follows, for example, from the structure of the rigidity matrix of a general mechanism, which has a block-diagonal form [24] with each block corresponding to an Assur Graph. There exists a systematic way to decompose any general parallel robot into its basic minimal components. Therefore, the singularity analysis of a general robot consists of two steps 1. Decomposition of the robot into its minimal components, using available algorithms (for example, the pebble game algorithm described in Ref. [24]). 2. Analysis of each minimal robot separately. Fig. 2 demonstrates the special singular properties of both 2D and 3D AGs. The topology of the 2D AG in Fig. 2a is called in the literature [32], the Triad; it is in a singular configuration, if and only if, the extensions of legs 1, 3 and 4 intersect at a

Fig. 1. An example of minimal parallel robot (a) and non-minimal parallel robot (b).

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Fig. 2. (a) 2D Triad and (b) spatial Triad in a singular configuration.

single point. Obviously, all the joints are mobile since link 2 rotates around (2,0), which is the absolute instant center of the triangle ABC, and the inner forces are in equilibrium, since links 1, 3 and 4 intersect at a single point. The same holds for the 3/6 SP shown in Fig. 2b: for example, consider the forces that exist in legs 1 and 2, F1 and F2 , respectively. These forces are directed along the direction of the corresponding legs. Due to the dual Aronhold-Kennedy theorem [34], their sum, F12 , should pass through their intersection point, i.e., through joint A. The dual Aronhold-Kennedy theorem [34] states then, that in the singular configuration, the three force sums F12 , F34 and F56 , intersect at the same point and lie in the same plane, thus, the inner forces in the legs are in equilibrium. Also, all the joints are mobile since the platform has rotational and translational velocities around the ISA, which in this case passes through the intersection point of the forces. In this condition, the velocity of each joint A, B and C is perpendicular to the corresponding leg, since the legs are rigid bodies. Gosselin and Angeles [5] introduced the classification of singularity into output, where ISAs of the platform exist in spite of zero actuation; input, where the actuation can be non-zero in spite of zero ISAs of the output platforms; and mixed, where both are present. In our approach, in order to check the output singularity, we lock the inputs and study the velocities of the resulting mechanism. For example, the piston actuators in spatial Triad (Fig. 2b) are presented in this analysis as solid bars. For the input singularity, we lock the output and free the actuators. Taking the same example (Fig. 2b) we have the triangle ABC frozen and every piston actuator is viewed as a mechanism that consists of two links connected by a slider. Consequently, while we specifically demonstrate finding output singularity in this paper, the same approach would work as well for input singularity. 4. The criterion for singularity and its application to the 6/6 SP This section introduces the novel singularity criterion for minimal parallel robots and its mathematical proof. The singularity criterion relies on the singular properties of Assur Graphs (Section 3). The criterion is subsequently demonstrated on the 6/6 SP. Next, it is shown how the criterion is applied in the planar case. 4.1. The general singularity criterion In a previous paper [19], we found the singular characterization of SP 6/6 relying on the assumption that in the singular configuration, there exist two lines that each intersects four legs. Although in all the known singular configurations of the 6/6 SP such two lines exist, it has not been yet proved that it is always possible to find such lines. Consequently, the new singular criterion does not require this assumption. The disadvantage of the proposed criterion is that it is not a geometric characterization; nevertheless, this time it is applicable to any topology of parallel robots. Theorem 1. (The singularity criterion): AG is in a singular position, if and only if for every link in it, after removing three arbitrary pairs of legs, one pair at a time, when no leg is removed three times, and after finding for each removal the center axis of the cylindroid of the possible ISAs of the remaining mechanism, the ISA of that link crosses and is perpendicular to the three center axes of the cylindroids. Proof. When the AG is in a singular position, each link possesses a unique ISA. If one arbitrary leg is removed, it becomes a 1DOF mechanism with the same ISA for each link. If we remove another arbitrary leg from the AG, it becomes a 2DOF mechanism. All the possible ISAs of a link of such a reduced mechanism belong to a cylindroid and are perpendicular to a central axis of this cylindroid. Therefore, the ISA of the above 1DOF mechanism belongs to this cylindroid. Repeating this process with another two pair of legs will give another cylindroid of the ISAs with its own central axis; and repeating it with the remaining third pair of legs gives a third cylindroid and its central axis. In an AG that is in a singular configuration,

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the ISA of the link belongs to each one of the three above mentioned cylindroids. Therefore, if the ISA exists, it should be a common normal to the three central axes. Conversely, suppose we have an AG in certain geometry and after performing the aforementioned steps, there is a common normal to three central axes of the cylindroids of the 2DOF mechanisms created by the deletion of two legs. Each pair of 2DOF mechanisms that are obtained by deleting two legs defines some ISA of the link. The three deletions assure that the ISAs of the mechanisms obtained by deleting different links are the same. Since no link is removed all three times, it is assured that all the links of the mechanism are compatible with the ISA. The properties of AG assure that this ISA is unique; therefore, there is a motion in the mechanism. Thus, the mechanism is in the singular position.  While the criterion is expressed in geometry, we can also represent it by different types of algebra, and here specifically we show it using Plücker coordinates and Grassmann-Cayley algebra. First, we determine a central axis of a cylindroid as a common normal to two of its possible ISAs, $1 and $2 . Each ISA can be obtained from the 1DOF mechanism derived from the 2DOF mechanism by freezing one of its DOFs. For the common normal to two screws, we have S = S1 × S2 , while its moment is found from the condition of reciprocality between it and each one of the two screws. Thus:

⎧ ⎫ ⎨S1 × S2 ⎬ CN1,2 = (S1 × S2 ) · S0 2 S1 − (S1 × S2 ) · S0 1 S2 . ⎩ × (S1 × S2 )⎭ ( S1 × S2 ) · ( S1 × S2 )

(1)

According to Lemma 2, perpendicularity and reciprocality assure that the common normal, as calculated above, intersects each one of the two screws. In order to check if three screws have a common normal, we need to check that their directions are in one plane and that the common perpendicular to the first two screws is reciprocal to the third. This results in the following two conditions:

( S1 × S2 ) · S3 = 0

(2)

( S1 × S2 ) · ( S0 1 ( S2 × S3 ) · ( S1 × S2 ) + S0 2 ( S3 × S1 ) · ( S1 × S2 ) + S0 3 ( S1 × S2 ) · ( S1 × S2 ) ) = 0

(3)

4.2. Application to the 6/6 SP This subsection applies the general criterion to the most general configuration of the 6/6 SP to demonstrate how it applies geometrically and algebraically. We remove two arbitrary legs, say 1 and 2. In order to find the center axis of the cylindroid of all the possible ISAs of the remaining four-leg platform, we find two possible ISAs. Their common normal is the center axis of the cylindroid since:

$1−2 = ω1 $1 + ω2 $2 ,

(4)

where screw $1−2 denotes any ISA of the SP and $1 and $2 are the two possible ISAs. The common normal of screws $1−2 , $1 and $2 is the center axis of the cylindroid of the four-leg platform that is obtained by removing the pair of leg lines 1–2. Each ISA is sought as a sum of two rotations. To do this, we draw two rotation axes as lines L1 and L2 , that cross three of the remaining four legs, as shown in Fig. 3a. Each one of these lines is obtained by choosing arbitrarily any point P on the first leg, and intersecting two planes that are built on point P and leg 2 or 3:



L1 = (P1 ∨ l2 ) ∧ (P1 ∨ l3 ) L2 = (P2 ∨ l2 ) ∧ (P2 ∨ l3 )

(5)

where P1 is the above descried point P chosen on leg l1 to build line L1 and P2 — to build line L2 . There is an infinite number of such lines, so in the same way there might be infinite possible ISAs. Since the lines of the rotations do not have to cross each other, the ISA that is their sum is, in general, not a line but a screw that we denote $1 :

$1 = ω1 L1 + ω2 L2

(6)

The projection of the platform velocity along each of the remaining four legs is equal to zero, since the legs are rigid bodies. This can be formulated as follows:

$1 ◦ li = 0, i = 1, . . . , 4

(7)

From Lemma 1, Eq. (7) is satisfied identically for the three leg lines that intersect lines L1 and L2 .

( ω1 L1 + ω2 L2 ) ◦ li = 0, i = 1, . . . , 3 Eq. (7) must also be valid for leg line l4 . Let line l4 be parallel to l4 and cross lines L1 and L2 , as shown in Fig. 3b. Its direction vector is the same as of l4 and its moment is found from the condition of the intersection between it and each one of the two rotation lines. Using these conditions, we obtain:



l4 =



s4 (s4 · S0 2 )S1 − (s4 · S0 1 )S2 . × s4 s4 · ( S1 × S2 )

(8)

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Fig. 3. A schematic explanation of the singular configuration of the 6/6 SP. The platform is rotated approximately 135° relative to the base. a. Lines L1 and L2 cross the three leg lines 1–3. b. Line l4 is parallel to l4 and crosses lines L1 and L2 . Plane π4,4 is built on l4 and l4 . c. The perpendicular plane π ⊥ and the line L⊥ . d. Screw $1 is the common perpendicular to C NL1 ,L2 and to L⊥ . e. Screw $2 , the cylindroid, and the common normal, CNI , to screws $1 and $2 (rotated for clarity). f. The singular criterion for the 6/6 SP: the ISA is perpendicular to the three common normals: CNI , CNII and CNIII .

As required by (7), $1 ◦ l4 = 0 and due to Lemma 1:

$1 ◦ l4 = (ω1 L1 + ω2 L2 ) ◦ l4 = 0

(9)

Thus, we derive the following equations:





$1 ◦ l4 − l4 = 0

(10)

Since l4 is parallel to l4 , s4 = s4 , then:



s4 s04

$1 ◦

$1 ◦

 0 





−

s4 s04





= $1 ◦

s04

0 − s04

 =0

(11)



Rl4 − Rl4 × s4

=0

(12)

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Fig. 4. The singularity criterion implemented on the 3/6 SP. a. The possible ISAs after removing each time a pair of legs, and the common normal of the ISAs. b. The known singular characterization of the 3/6 SP — four planes intersect at a single point. c. The singular characterization for the 3/6 SP: the ISA is perpendicular to the three common normals CNI , CNII and CNIII . In addition to the indication whether the 3/6 SP is in a singular position, the presented method also determines the ISA of the platform in this position.

where Rl4 and Rl  are some points on l4 and l4 , respectively. 4

A plane π4,4 , is built on l4 and l4 . To describe algebraically such a plane, it is sufficient to take 3 points: a point at infinity in the direction of both lines, and one arbitrary point from each line; for example, the point that is obtained through the moment and the direction of the line:

π4 , 4 

⎫ s



s ⎧s0 4 − s0 4 s0 s0 4 ⎨  ⎬ s4 4 4 × 4 = |s4 | |s4 | ∨ |s4 | × |s4 | ∨ |s4 | = s4 · s0 4 × s0 4 ⎩ ⎭ 1 0 1 2

(13)

s4

Rl4 − Rl  and s4 lie in, or are parallel to, plane π4,4 , thus (Rl4 − Rl  ) × s4 = N4,4 , where N4,4 is normal to plane π4,4 . 4

After substituting the normal to Eq. (11) we obtain:

   S1 S01

0 N4,4

◦

4



= S 1 · N4,4 = 0

(14)

The latter multiplication is equal to zero, if and only if:

$1 ⊥ N4,4 π4,4

(15)

Which means that the screw is parallel to plane π4,4 or perpendicular to its normal N4,4 . Let plane π ⊥ be perpendicular to plane π4,4 and parallel to the common normal of lines L1 and L2 , C NL1 ,L2 (Fig. 3c). Using Plücker coordinates:



π⊥ =

P·





s0 4 − s0 4 × (S1 × S2 )







s0 4 − s0 4 × (S1 × S2 )

(16)

where P is a vector to an arbitrary point in π4,4 . From Eqs. (15) and (16) it follows that screw $1 is perpendicular to plane π ⊥ . Let line L⊥ lie on the perpendicular plane, thus being perpendicular to the screw, and cross lines L1 and L2 , as shown in Fig. 4c. Using the notation of Grassmann-Cayley algebra:

L⊥ =



 π ⊥ ∧ L1 ∨ π ⊥ ∧ L2 .

(17)

The reciprocal product of L⊥ with the screw $1 is identically zero since L⊥ crosses lines L1 and L2 .

( ω1 L1 + ω2 L2 ) ◦ L⊥ = 0

(18)

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Therefore, according to Lemma 2, screw $1 crosses line L⊥ , thus, screw $1 is well defined and is a common perpendicular to the common normal of lines L1 and L2 , C NL1 ,L2 , and to L⊥ , as shown in Fig. 3d. When the screw is a common perpendicular to the common normal of lines L1 and L2 and to line L⊥ , this screw satisfies the necessary conditions (7) of the ISA of the four-leg platform. Another ISA, $2 , is found by repeating the above process with two different lines. According to Eqs. (4) and (7), the reciprocal product of $1−2 with any of leg lines 1–4 is equal to zero. Due to the generalized Aronhold-Kennedy theorem, any ISA $1−2 of the platform obtained from Eq. (4) should be perpendicular to the common normal of screws $1 and $2 , and cross it. This common normal is the center axis of the cylindroid, designated by CNI , as shown in Fig. 3e. This is precisely the cylindroid of a system of two screws [35]. It is analogous to the 2D case, where for the system of 2DOF, all the possible instant centers of any link lie on a line [36]. Note, that if there exist two lines that cross the four legs, as in Ref. [19], these lines are the two ISAs that are sought after through the described process. We repeat this process by removing another two legs, let us say 3 and 4, for which we draw twice two lines that define another two screws. CNII is the common normal of the latter two screws, which is the center axis of the cylindroid of this 2DOF platform. The ISA of the original platform is the common perpendicular to the two centers of axes of the two cylindroids, i.e., common perpendicular to CNI and CNII . Now, we come to the singular criterion and the singular properties that exist only in AGs. We repeat for the third time the process of removing two legs, drawing twice two lines and finding another center axis of another cylindroid, i.e., CNIII . According to Theorem 1, the 6/6 Stewart Platform is in a singular configuration if and only if the three ISAs: $1−2 , $3−4 and $5−6 , obtained for each deleted pair of legs, are actually the same screw. In other words, the SP is in a singular position if the ISA is perpendicular to the three center axes of the three cylindroids, i.e., to CNI , CNII and CNIII , as shown in Fig. 3f. This condition determines that the mechanism is in a special position, where inner forces exist in all the elements. Consequently, according to the special properties of AGs, (referred to in Section 3) the mechanism is in a singular configuration. 4.3. Application to planar mechanisms Planar mechanisms can be viewed as a subset of spatial mechanisms where, all the ISAs are perpendicular to the xyplane. Thus, considering such a mechanism as planar, all the ISAs are viewed as instant centers of rotations. The AronholdKennedy theorem is reduced to its 2D version [37], which states that three relative instant centers of three bodies lie on a straight line. The center axis of the cylindroid of a 2DOF mechanism that is a common normal to two possible ISAs becomes a line that connects two possible instant centers. Finally, the crossing of the ISA of the three center axes of the cylindroids becomes an intersection of three such lines. Let us take as an example the Triad (Fig. 2a). First, we remove legs 3 and 4. The locus of the instant centers of rotation of link 2 of the remaining mechanism coincides with leg 1, since the two possible instant centers are joint A and the joint that connects leg 1 to the ground. The same logic gives leg 3 as the locus of the centers of rotation after removal of legs 1 and 4, and leg 4 after removal of legs 1 and 3. Therefore, the Triad is in a singular configuration, if and only if, the extensions of legs 1, 3 and 4 intersect at a single point. 5. Comparison to a known singularity result In this section we apply the singularity criterion to a 3D bar-and-joint triad known as the 3/6 SP. We compare the outcome of the proposed criterion with the singularity result that appears in the literature in the context of the 3/6 SP singularity. Merlet [9] studied the singularity of six-DOF 3/6 SP based on Grassmann line geometry. Downing et al. [13], followed by Ben-Horin and Shoham [10], proved by relying on Grassmann-Cayley algebra, that all the singular positions of the 3/6 SP are delineated to be the intersection of four planes at a single point. They showed that all the singular configurations of Merlet are particular cases of this condition. This geometric condition consists of four planes, defined by the actuator lines and the position of the spherical joints, which intersect at a single point. We derive the singularity of the 3/6 SP by applying the proposed singularity criterion; the result is then compared to the above characterization reported in the literature. It should be noted, that the criterion presented here gives not only the indication whether the 3/6 SP is in the singular position, but also determines the line of the ISA of the platform in this position. Let πi be the plane formed by the two leg lines meeting at joint Ai . We remove two pair of legs from the SP, say 1 and 2, and draw two screws, $1 and $2 , that cross the four remaining legs. The two planes π2 and π3 , share an intersection line; let us define $2 as the line A2 A3 and $1 as the intersection line between the planes containing π2 and π3 , as shown in Fig. 4a. In this way, the reciprocal product of each of the screws with each of the leg lines is equal to zero, since these lines cross the four remaining legs. According to Eq. (7), these two lines are possible ISAs of the four-leg platform. Due to the generalized Aronhold-Kennedy theorem, any ISA of the platform obtained from equation (4) should be perpendicular to the common normal of screws $1 and $2 and cross it. This common normal is the center axis of the cylindroid, designated by CNI , and is given in Fig. 4a. We repeat this process by removing another pair of legs, e.g., 3 and 4, for which we draw twice two screws that define another common normal CNII , shown in Fig. 4a, which is the center axis of another cylindroid of this platform. The ISA of the platform is the common perpendicular to the two centers of axes of the two cylindroids,

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i.e., common perpendicular to CNI and CNII . Now, we come to the singular criterion and the singular properties that exist in AGs. We repeat for the third time the process of removing a pair of legs, drawing twice two screws and finding another center axis of another cylindroid, i.e., CNIII , as shown in Fig. 4a. The 3/6 SP is in a singular position if and only if ISA $t of the SP is perpendicular to CNI , CNII and CNIII , as defined in the proposed singular criterion, (Section 4) as shown in Fig. 4b. We are going to show that in this condition the three planes formed by the pairs of leg lines meet at a single point on the platform plane, as shown in Fig. 4c. As mentioned, this is a known singular characterization of the 3/6 SP [13,10]. In effect, the ISA is the linear combination of rotations around line A2 A3 (designated as $2 ) and around the intersection line of planes π2 and π3 (designated as $1 ). Therefore, the velocity of any point on line $1 is perpendicular to line $2 ; thus, the velocity of the point of intersection of the platform plane and planes π2 and π3 is perpendicular to line A2 A3 , and therefore to the platform plane. The same is true for the point of intersection of the platform plane and the planes π1 and π2 . And the same is true for the point of intersection of the platform plane and the planes π1 and π3 . Therefore, there are three points on the platform in which the velocity is perpendicular to the platform plane. But since the platform is a rigid body, there can be only one point that has a given direction of velocity, thus these three points are the same point. 6. Examples of multi-platform robot 6.1. Two-platform robot with 7 legs and a hinge While Section 5 compared the results of applying the new criterion to results available in the literature, this section demonstrates how the criterion is used to determine a singular position of a robot whose singular characterization does not appear in the literature. The result is verified numerically by using the Jacobian of the robot in the singular position. We consider a parallel robot that consists of two platforms, connected with a hinge L and 7 legs: four legs are connected to platform I pairwise at points B1 and B2 and three legs are connected to platform II at points B3 , B4 and B5 , as shown in Fig. 5a. Specifically, line 1 starts at point A1 and goes to point B1 , line 2 goes from point A2 to B1 , line 3 — A3 to B2 , line 4 — A4 to B2 , line 5 — A5 to B3 , line 6 — A6 to B4 and line 7 — A7 to B5 . Line L goes from P to Q (Fig. 5b). This parallel robot is an example of an Assur Graph, since the removal of any element causes a motion in all the other elements. It has zero DOF, since it consists of two bodies, 7 legs that provide a single constraint each, and a hinge that provides 5 constraints:

2×6−5−7=0 Given that the robot is an Assur Graph, it is sufficient to check only one of its platforms. We will analyze its singularity by the method proposed in this article: for each removal of one of the following pair of legs: 5–6, 6–7, 5–7, we find the center of the cylindroid of the ISAs of platform II. The platform is in a singular position if the three central axes of the cylindroids have a common perpendicular. In each case of legs removal, we have a 2DOF mechanism. The ISA of platform I can be presented as a sum of two rotations:

$I = ω1 L1 + ω2 L2

(19)

where L1 is the line of intersection of the planes defined by legs 1–2 and 3–4 and L2 is the line connecting joints B1 and B2 . Body I is connected by hinge L with body II, while body II is connected with the basis through a leg with two spherical joints. This kinematic chain is presented schematically in Fig. 6. O and C denote respectively, joints: in case of removing legs 5–6 — A7 and B5 ; in case of removing legs 6–7 — A5 and B3 ; and in case of removing legs 5–7 — A6 and B4 . Two possible ISAs are needed to find the central axis of the cylindroid of II. We find them by setting alternatingly ω1 = 0 and ω2 = 0. In each case the kinematic chain transforms, as shown in Fig. 7a.

Fig. 5. Multi-platform robot. a. Scheme. b. General view.

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Fig. 6. Kinematic chain of the platform when two legs are removed.

Fig. 7. a. Kinematic scheme for determining the ISA of body II when legs 6 and 7 are deleted. b. Determining ISA $1 when rotation around L2 is fixed.

The ISA of platform II is a sum of rotations around L1 and L from one side, and of rotations around the axes that pass through O (let us call it LO ) and through C (will be called LC ) from the other side. The Aronhold-Kennedy theorem states that the ISA is perpendicular to the common perpendicular of L1 and L (denoted as C NL,L1 in Fig. 7b). On the other hand, since the ISA is a sum of rotations around LO and LC , it is reciprocal to l5 :

$II ◦ l5 = (ωO LO + ωC LC ) ◦ l5 = 0

(20)

Choosing l5 so that it is parallel to l5 and crosses L1 and L, we have that the ISA is identically reciprocal to l5 : $ ◦ l  = ω L + ωL ◦ l  = 0 II

5

(

1 1

)

5

(21)

Therefore, by the same logic in Eqs. (9)–(15), the ISA is parallel to plane π5,5 that is built on l5 and l5 . Therefore, to determine the position of the ISA, we build plane π ⊥ that is parallel to the common perpendicular of L1 and L (C NL,L1 ) and perpendicular to π5,5 . We build a line L⊥ through the points of intersection between π ⊥ and lines L1 and L. The ISA is the common perpendicular to this line and C NL,L1 . This ISA will be denoted as $1 (Figs. 7b and 8a). By choosing ω2 = 0 and applying the same process, we obtain $2 (Fig. 8a). The central axis of cylindroid CN5 is a common perpendicular to $1 and $2 , as shown on Fig. 8a. As described above, these operations are repeated on the mechanism with deleted legs 5–6 and 5–7. Finally, it is checked whether the three central axes of the cylindroids share a common perpendicular (Fig. 8b). We will take a particular singular position, and verify that the Jacobian vanishes and the present criterion is valid. The Jacobian is obtained from the following equations:

$I ◦ li = 0, i = 1 . . . 4

(22)

$II ◦ li = 0, i = 5 . . . 7

(23)

ωII $II − ωI $I = ωL

(24)

The Jacobian of the second platform is:

l1 JII = L ◦ l1

l2 L ◦ l2

l3 L ◦ l3

l4 L ◦ l4

l5 0

l6 0

l7 0

 (25)

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Fig. 8. Analysis of the platform with the geometry obtained from the Jacobian. a. Determining the cylindroid central axis when legs 6 and 7 deleted. b. Determining the three central axes and obtaining ISA $II .

For the numerical example, we take the following coordinates for the basis joints:



A1 =





−9 −2 , A2 = 0



−6 −5 , A3 = 0





And the following for the platform joints:



B1 =





−5 −2 , B2 = 5



−5 4 , B3 = 7



−7 5 , A4 = 0



3 3 , B4 = 4



−2 8 , A5 = 0





4 −1 , B5 = 5





6 6 , A6 = 0



3 −4 , P = 5





x 3 , A7 = 0



1 −2 , Q = 7





7 −3 0



0 0 4

The value of x will be determined through the Jacobian. The Jacobian of this mechanism is:

  4   0  5  J = −10  45  8   44

1 3 5 − 25 30 − 13 104

2 −1 7 35 49 −3 60

−3 −4 7 56 14 32 − 84

−3 −3 4 24 − 24 0 0

4−x −4 5 15 − 5x − x − 12 0



−4  −1 5  − 15 = 0 − 35  − 19 0 

which gives x = 5.791. With this value, we now have a singular position of the parallel robot, due to its vanishing Jacobian. The analysis according to the method proposed in this article is presented in Fig. 8. The following are some numerical results for this example. In order to spare the reader large integer numbers, the Plücker coordinates are normalized:

L=P∨Q =

⎧ ⎫ −0.27⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.54 ⎪ ⎬

⎧ ⎫ 0.57 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.085 ⎪ ⎬

⎧ ⎫ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.949 ⎪ ⎬

−2.14⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩−1.07⎪ ⎭

1.33 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 10.32 ⎪ ⎭

−5.38⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1.58 ⎪ ⎭

−0.8

, L1 = ( A1 ∨ B1 ∨ A2 ) ∧ ( A3 ∨ B2 ∨ A4 ) =

0.82

−1.99

0

, L2 = B1 ∨ B2 =

0.316

−4.74

where the calculations are the standard procedures of Grassmann-Cayley algebra. Using Eqs. (1) and (8), we can calculate the lines in Fig. 7b:

CNL,L1 =

⎧0.78 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −0.37⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.51⎪ ⎬

⎧−0.51⎫ ⎪ ⎪ ⎪ ⎪−0.51⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨0.69 ⎪ ⎬

⎪ 2.04 ⎪ ⎪ ⎪ ⎪ 3.89 ⎪ ⎪ ⎩

⎪ 1.25 ⎪ ⎪ ⎪ ⎪ 8.43 ⎪ ⎪ ⎩

0.32

, l5 =

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

7.26

.

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

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The plane on two parallel lines is built using Eq. (13) and the plane perpendicular to the given plane and parallel to the given line is:





S×N −d

π⊥ =

(26)

where S is the direction of the line, N is the normal to the plane, and d is any arbitrary number. Thus, in the above example in Fig. 7b (in normalized Plücker coordinates):

π5 , 5  =

⎧ ⎫ 0.19 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.85⎪ ⎬

,

⎪ −0.49⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭

π⊥ =

3.93

⎧ ⎫ 0.36 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.41⎪ ⎬ ⎪ 0.84 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 0.32

where it was chosen that plane π ⊥ will pass through A5 . Line L⊥ through the points of intersection between π ⊥ and lines L1 and L is:

L⊥ =



 π ⊥ ∧ L ∨ π ⊥ ∧ L1 =

⎧ ⎫ −0.92⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.011 ⎪ ⎬ 0.39

.

0.77 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩−0.53⎪ ⎭ 1.82

The first ISA in Fig. 7b is the common normal of the above line and C NL,L1 :

$1 =

⎧0.35 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −0 . 41 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨0.84 ⎪ ⎬ ⎪ 2.42 ⎪ ⎪ ⎪ ⎪ 2.96 ⎪ ⎪ ⎩ 0.41

.

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

The same procedure with fixing rotation around L1 gives the second ISA in Fig. 8a, and the central axis of the cylindroid of the mechanism with legs 6 and 7 removed:

$2 =

⎧−0.28 ⎪ ⎪ ⎪ ⎪ 0.24 ⎪ ⎪ ⎪ ⎨−0.93

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

⎧0.87 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0.48 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.13⎪ ⎬

⎪ −0.92 ⎪ ⎪ ⎪ ⎪ −1.21 ⎪ ⎪ ⎩

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎪ −4.72⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 9 . 17 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭

, CN5 =

−0.041

.

2.30

Analogously, the central axes of cylindroids with legs 5, 7 and 5, 6 removed, respectively, are obtained as:

C N6 =

⎧ 0.85 ⎪ ⎪ ⎪ ⎪ ⎨ 0.52

⎫ ⎪ ⎪ ⎪ ⎪ ⎬

⎧ ⎫ −0.81⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.53⎪ ⎬

−6.89⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 10.03 ⎪ ⎭

9.08 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩−9.66⎪ ⎭

0.098

, C N7 =

6.52

−0.26

.

−8.72

It can be validated using the conditions (2) and (3) that the three central axes share a common perpendicular. The ISA of platform II is the common normal of CN5 and CN6 :

$II =

⎧−0.50⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0.85 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.18 ⎪ ⎬

.

⎪ −8.54⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−5.91⎪ ⎪ ⎪ ⎪ ⎩ ⎭ −4.14

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Fig. 9. 1TR3 robot. a. Scheme. b. Geometry.

Fig. 10. Removal of link 2 and determination of central axis CN2 .

It can also be validated that $II ◦ CN7 = 0. It should be noted that the geometric algorithm gives us the line vectors, without the pitches. For example, it can be validated that the Plücker condition S · S0 = 0 is valid for screw $II above. In order to determine the pitch of the ISA $II , we need to maintain a reciprocality condition between this ISA and one of the legs (Eq. (23)), for example, l5 :

$II ◦ l5 = S · S0 5 + S0 · S5 + hII S · S5 = 0 that gives hII = −3.06. We thus demonstrated that the new criterion offered in this paper can determine whether the current position of this robot platform is singular and the result also provides the direction of the ISA. 6.2. 1TR3 two-platform robot Here we take a 1TR3 robot described by Song et al. [25]. The robot consists of three platforms, where the third platform is connected by a helical joint with the first platform and with a hinge to the second platform. For the sake of this example, we omitted the third platform and connected the two basic platforms by a cylindrical pair; this simplification does not affect the singularity analysis. Each one of the two platforms is connected to a link by a spherical joint and the link is connected to the ground by a hinge. The scheme of this robot is shown in Fig. 9a and its geometry in Fig. 9b. The spherical joints of links 1–4 are denoted by A to D respectively. The cylindrical pair is denoted as $. The points A to D denote the connection of the links 1–4 with the corresponding hinge axes which are lying in the basis plane. The links are connected with the axes by a right angle. Every link of this robot can be replaced by two legs with one end on the corresponding spherical joint and the other end on the corresponding hinge axis. Therefore, removing one link in this robot is equivalent to removing two legs. We analyze platform II by removing links 2, 3 and 4, sequentially. First, we remove link 2 and find the center axis of the cylindroid of the ISAs of platform II, CN2 , as the common normal between two possible ISAs. To find those two, we notice first, that in this configuration (Fig. 10), the ISA of platform II does not depend on platform I. The first such ISA is a line L3 , that is a line through C and D, since platform II can rotate around an axis defined by those two points. To find the second possible ISA, we note that the velocity of joint C is perpendicular to the plane that passes through the hinge axis of link 3 and point C itself. Similarly, the velocity of joint D is perpendicular to the plane that passes through the hinge axis of link 4 and point D. Thus, the second possible ISA is line L4 , that is the intersection of the two above defined planes (Fig. 10).

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Fig. 11. a. Kinematic chain when link 3 or 4 is removed. b. Kinematic scheme for determining the possible ISA of body II when link 3 and 4 is removed.

Fig. 12. a. Construction when link 3 is removed and ω2 = 0. b. Determination of ISA $1 . c. Determination of the central axis CN3 . d. The three central axes and ISA $II .

Removing link 3 or link 4 creates a mechanism shown schematically in Fig. 11a. L2 is a line through A, B, and L1 is the intersection of the planes defined by links 1 and 2. X denotes joint C or D and l denotes hinge 3 or 4, respectively. The ISA of platform I is a sum of two rotations as per Eq. (19). We find the two possible ISAs by setting alternatingly ω1 = 0 and ω2 = 0. In this case the scheme in Fig. 11a transforms into the scheme in Fig. 11b. The ISA of platform II, $II is a sum of rotations around L1 and $ from one side, and of rotations around l and the axis that passes through X (denoted further LX ) from the other side. The Aronhold-Kennedy theorem states that the ISA is perpendicular to the common perpendicular of L1 and $ (denoted as CN$,L1 in Fig. 12a). We build line L⊥ to be perpendicular and reciprocal to $II . For this purpose, we build line L that passes through point X, crosses line l and is perpendicular to the direction of screw $. To build such a line it is sufficient to build a plane through X that is perpendicular to $ and the required line passes through the intersection point of this plane and hinge axis l. Due to Lemma 1:

$II ◦ L = (ωl l + ωX LX ) ◦ L = 0

(27)

Choosing L so that it is parallel to L and crosses L1 and $, we have that the ISA is identically reciprocal to L due to Lemma 2:

$II ◦ L =



ω1 L1 + ω $ ◦ L = 0

(28)

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Therefore, by the same logic in Eqs. (9)–(15), the ISA is parallel to plane πL,L that is built on L and L . Therefore, to determine the position of the ISA, we build plane π ⊥ that is parallel to the common perpendicular of L1 and $ (CN$,L1 ) and perpendicular to πL,L , and passes through point X. We build line L⊥ through the points of intersection between π ⊥ and lines LX (point X) and l. Due to Lemma 1:

$II ◦ L⊥ = (ωl l + ωX LX ) ◦ L⊥ = 0

(29)

These steps are illustrated in Fig. 12a. The ISA is the common perpendicular to this line and CN$,L1 . This ISA will be denoted as $1 (Fig. 12b). By choosing ω2 = 0 and applying the same process, we obtain $2 (Fig. 12c). Performing these operations when link 3 is removed, the central axis of cylindroid CN3 is obtained as a common perpendicular to $1 and $2 , as shown on Fig. 12c. The same operations performed when link 4 is deleted give the central axis of cylindroid CN4 . Finally, it is checked whether the three central axes of the cylindroids share a common perpendicular (Fig. 12d). We verify the criterion against the result obtained by using a Jacobian in a particular singular position. In order to obtain the Jacobian, we use two “legs” equivalent to each link — one “leg” connects the corresponding joint with the hinge axis being perpendicular to it, (denoted li,⊥ , where i is the number of link) and the second “leg” passes through the joint and is parallel to the corresponding hinge (denoted li, ). We have:



$I ◦ li,⊥ = 0

, i = 1...2

$I ◦ li, = 0



$II ◦ li,⊥ = 0 $II ◦ li, = 0

(30)

, i = 3...4

(31)

  S S0

ωII $II − ωI $I = ω$ = ω

  +h

0 S

  =ω

S S0

  +v

0 S

  = ω L$ + v

0 S

(32)

The Jacobian of the second platform is:

⎡

l1,⊥

JII = ⎣ L$ ◦ l1,⊥ S 1,⊥ · S

l1,

l2,⊥

l2,

l3,⊥

l3,

l4,⊥

l4,

L$ ◦ l1,

L$ ◦ l2,⊥

L$ ◦ l2,

0

0

0

0

S 1, · S

S 2,⊥ · S

S 2, · S

0

0

0

0

⎤ ⎦

(33)

For the numerical example, we take the following coordinates for the basis joints:

A=



4 −4 , B = 5





−4 4 , C= 3





−2.5 −2.5 , D = 1.5





3.5 3.5 4.5

and for the connections of the links to the correspondent hinge axes:

A =

⎧ ⎫ ⎨5.454⎬

⎧ ⎫ ⎨−8⎬

⎩

⎩

−7 0

, B =

⎭

7 0

, C =

⎭

⎧ ⎫ ⎨−3⎬

⎧ ⎫ ⎨4⎬

⎩

⎩ ⎭

−4 , D = 0

⎭

5 0

The cylindrical pair goes through the midpoint between A and B and the midpoint between C and D. Therefore the line of cylindrical pair in the Plücker coordinates is:

L$ =

⎧−0.5⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −0.5⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨1 ⎪ ⎬ ⎪ 2 ⎪ ⎪ ⎪ ⎪−2 ⎪ ⎪ ⎩ 0

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

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Thus, the Jacobian of the second platform is:

 −1.454   3   5   −35 JII =  −27.269  6.185   28.412 

−3 −1.454 0 7.269 −15 −17.815 −17.042 2.227

4.227

4 −3 3 21 24 −4 −12.5 2.5

3 4 0 −12 9 −28 −28.5 −3.5

0.5 1.5 1.5 −6 4.5 −2.5 0 0

−1.5 0.5 0 −0.75 −2.25 −5 0 0

−0.5 −1.5 4.5 22.5 −18 −3.5 0 0



1.5  −0.5 0   2.25  =0 6.75  −7  0   0

Performing the aforementioned steps and implementing the criterion, gives the following center axes:

C N2 =

⎧ ⎫ 0.116 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.349 ⎬

⎧ ⎫ 0.0862 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.373 ⎪ ⎬

⎧ ⎫ 0.00846⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0.100 ⎪ ⎬

−4.929⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 3.872 ⎪ ⎭

−0.528⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −2 ⎪ ⎩ .266⎪ ⎭

6.207 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 12 ⎪ ⎩ .851 ⎪ ⎭

−0.930

, C N3 =

0.836

0.924

0.963

, C N4 =

0.995

−1.350

It can be validated again using conditions (2) and (3) that the three central axes share a common perpendicular. The ISA of platform II can be calculated as the common normal of CN2 and CN3 :

$II =

⎧−0.669 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0.188 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨−0.0132⎪ ⎬ ⎪ −1.452 ⎪ ⎪ ⎪ ⎪ −5.017 ⎪ ⎪ ⎩ 2.250

.

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

As in the previous section, we have obtained the line of the ISA. To calculate the pitch we can use, for example, Eq. (31). The pitch obtained is hII = −0.169. 7. Conclusions We presented a novel singularity criterion for spatial parallel mechanisms. This criterion uses the dual form of the wellknown Aronhold-Kennedy theorem in statics and the singular properties of Assur Graphs. Its main difference from the methods used in the literature is that it gives a geometric meaning to the singularity condition. The criterion is validated using results that appear in the literature, as shown in Section 5. An example of the 6/6 SP in a most general configuration shows that it is an effective methodology to deal with the singularity problem of spatial constrained parallel mechanisms. Further examples of finding the singularity of complex constrained parallel robots demonstrate the general applicability of the criterion to minimal parallel robots. In a previous paper [19], we found the singular characterization of the 6/6 SP, relying on the conjecture that in the singular configuration there exist two lines that intersect four legs. Although in all the known singular configurations such two lines exist, as yet it has not been proven. The singularity criterion introduced in this paper does not require such two lines. The limitation of the proposed method is that it is only a criterion for singularity and not a characterization, but this time it applies to all 3D minimal parallel robots (Assur Graphs). Since there is a systematic way of constructing new Assur Graphs and mechanisms based on AGs [38], the new criterion can be applied to many new types of parallel robots that will be developed in the future. And since any robot can be decomposed into its minimal components [24], the criterion is applicable to any complex mechanism. Finally, this paper dealt with the output singularity of minimal parallel robots; however, the methodology presented here allows determination of the input singularity in the same way. In this paper, we did not cover the issue of computational complexity. We presume, however, that the presented method that consists of a sequence of direct steps allows for faster computation than methods that are based on inversion of matrices or calculation of determinants. We leave this topic to future study. Acknowledgment This research was supported by the Ministry of Science, Technology, & Space, Israel and the Russian Foundation for Basic Research, the Russian Federation. References [1] E.F. Fichter, E.D. McDowell, A novel design for a robot arm, in: Computer Technology Conference, New York, 1980, pp. 250–256.

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