Conditions of regularizability

59 4. Regularization of the inverse positioning problem

that are all orthogonal to the vector p(θ) −q(θ), so they are from the plane orthogonal top(θ)−q(θ). Since the image space ofJ_V^e is spanned by the vectors (4.92-4.94), and these vectors are from a plane, the rank of the matrix is at most two.

Spherical manipulators have another unique property:

Proposition 12. Manipulators have joint axes intersecting at the same point if and only if there is a point on the last segment for whichJ_V^e = 0.

Proof. If the joint axes of the manipulators intersect at some pointq(θ) in the joint configurationθ∈ R³, while the end effector position isp(θ), then the linear velocity generators are defined by (4.92)-(4.94). Since q(θ) is on the last joint axis as well, it is on the last segment, so by choosingp(θ) =q(θ), the generators in (4.92)-(4.94) are all zero, soJ_V^e = 0.

Suppose that for a point on the last segment J_V^e = 0 in the joint configurationθ ∈ R³. Let this point bes(θ). Since J_V^e = 0, this implies that

v^e₁(θ) = ω^e₁(θ)×(s(θ)−q1(θ)) = 0 (4.95) v^e₂(θ) = ω^e₂(θ)×(s(θ)−q2(θ)) = 0 (4.96) v^e₃(θ) = ω^e₃(θ)×(s(θ)−q₃(θ)) = 0. (4.97) So for every jointi∈ {1,2,3}, eitherω_i^e(θ)is parallel to(s(θ)−q_i(θ))or s(θ) =q_i(θ). Ifs(θ) =q_i(θ), that means thats(θ)is on theith joint axis.

Ifω_i^e(θ)is parallel to(s(θ)−q_i(θ))that also means thats(θ)is on theith joint axis, sinceq_i(θ)is on theith joint axis. Sos(θ)is a point on all joint axes, thus the joint axes intersect in the same point.

60 4. Regularization of the inverse positioning problem

it can be written as

J^reg=π_V Ad_(I,−γr)J_e, (4.99) whereπ_V is the projector matrix

π_V =



 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0



. (4.100)

Since the rank of J^reg can not be greater than the minimal rank of the matrices in the right-hand side of (4.99), and rankπV = 3, while rank Ad_(I,−γr) = 6, sinceAdis an automorphism (that yields it is invert-ible), andrankJ^e≤3, the rank ofJ^regis

rankJ^reg≤min



rankπ_V

| {z }

=3

,Ad_(I,−γr)

| {z }

=6

,rankJ^e



= rankJ^e. (4.101)

SorankJ^reg = 3implies thatrankJ^e = 3.

Recall that by Proposition 5 the ranks of the end effector Jacobian, spatial manipulator Jacobian and body manipulator Jacobian are iden-tical. This implies that if the task Jacobian is regularizable, thenrankJ^s= 3and rankJ^b = 3as well.

SinceJ^reg = Ad_(I,−γr)J^e

V, and(I,−γr) is a translational genera-tor, the regularized Jacobian has the columns

J^reg = v₁^e+γω^e₁×r v^e₂+γω^e₂×r v^e₃+γω₃^e×r

= v₁^e v₂^e v₃^e

+ γr×ω^e₁ γr×ω₂^e γr×ω₃^e (4.102) and by introducing the following notation

γJ_Ω^e ×r= γω^e₁×r γω₂^e×r γω₃^e×r

, (4.103)

J^reg takes the short form

J^reg=J_V^e +γJ_Ω^e ×r. (4.104) Even if the task Jacobian is regularizable, it is important to know how to get the regularization vectorr. The following theorem constricts the search space for the regularization vector, by stating that the reg-ularization vector must be the linear combination of the linear velocity generators, i.e. it must lie in the image space ofJ_V^e.

Theorem 7. Ifr∈R³ is a regularization vector, thenr∈RanJ_V^e.

61 4. Regularization of the inverse positioning problem

Proof. Since the regularized Jacobian is

J^reg=J_V^e +γJ_Ω^e ×r, (4.105) its range space is

RanJ^reg⊆RanJ_V^e ∪Ran (J_Ω^e ×r) (4.106) where ∪ is the subspace union. Suppose indirectly, that r /∈ RanJ_V^e. Note that this impliesr 6= 0, since the zero vector is element of every subspace. SinceJ_Ω^e ×r is orthogonal tor,r /∈Ran (J_Ω^e ×r), however be-cause of (4.106) it follows thatr /∈RanJ^reg. This implies thatJ^regis not full rank, that contradicts with the statement thatris a regularization vector.

Theorem 6 showed that a necessary condition for regularizability is thatJ^e is full rank. This implies that the kernels of the Jacobians J_V^e andJ_Ω^e has only one common element, that is the zero vector:

Lemma 2. IfrankJ^e = 3, then there exists no nonzeron∈R³, such that n∈KerJ_V^e andn∈KerJ_Ω^e.

Proof. Suppose indirectly, that n 6= 0 ∈ R³, and n ∈ KerJ_V^e and n ∈ KerJ^eΩ. Since the end effector Jacobian is

J^e= J_V^e

J_Ω^e

, (4.107)

nis in the Kernel ofJ^e, since J^en=

J_V^e J_Ω^e

n=

J_V^en J_Ω^en

= 0, (4.108)

which implies thatJ^eis not full rank, that is a contradiction.

Lemma 3. LetQ6= 0be a3×3symmetric real matrix with zero elements in the diagonal, andS be a subspace ofR³ with dimS ≤2, and λ∈R³. Then the solution of the quadratic equation< Qλ, λ >= 0restricted toS is one of the following:

1. A line passing through the origin.

2. The origin.

3. Two lines passing through the origin.

62 4. Regularization of the inverse positioning problem

Proof. LetP_Sbe the orthogonal projection to the subspaceS, thendimS ≤ 2 implies rankP_S ≤ 2. The variables λ restricted toS areP_Sλ, so the equation< Qλ, λ >= 0restricted toScan be written as

< QPSλ, PSλ >=< P_S^⊤QPSλ, λ >= 0. (4.109) Introducing Q_S = P_S^⊤QP_S, the quadratic equation restricted to S be-comes

< Q_Sλ, λ >= 0. (4.110) SinceQ6= 0 is symmetric, with zero diagonals, its rank is at least two, while rankP_S ≤ 2, so by the rank nullity theorem and the property of product rank, 1 ≤ rankQ_S ≤ 2. Moreover, Q_S is also a symmetric real matrix, so its eigenvalues are real, and it is diagonalizable, i.e. it can be written in the form

QS =U





κ₁ 0 0 0 κ₂ 0

0 0 0



U^⊤ (4.111)

withU U^⊤=I, andκ₁, κ₂ ∈R. Substituting this form into the quadratic equation results in

λ^⊤U



 κ₁ 0 0 0 κ₂ 0

0 0 0



U^⊤λ= 0. (4.112)

Introducing the rotated variablesλ˜ = (˜λ₁,λ˜₂,˜λ₃)^⊤ defined as˜λ=U^⊤λ, the quadratic equation becomes

λ˜₁ λ˜₂ ˜λ₃



 κ₁ 0 0 0 κ₂ 0

0 0 0







 λ˜₁ λ˜₂ λ˜₃



=κ₁λ˜²₁+κ₂λ˜²₂= 0. (4.113)

Note that since the component ˜λ₃ ∈/ S, the subspace S after the trans-formationU^⊤ is spanned by˜λ₁ and˜λ₂.

Suppose that κ₂ = 0. In this caseκ₁ 6= 0, because this would imply Q = 0, that is a contradiction. The quadratic equation (4.113) reduces toκ₁˜λ²₁ = 0, and the solution is ˜λ₁ = 0, ˜λ₂ ∈ R, that is a line passing through the origin.

Suppose that κ1 = 0. In this caseκ2 6= 0, because this would imply Q = 0, that is a contradiction. The quadratic equation (4.113) reduces toκ₂˜λ²₂ = 0, and the solution is ˜λ₁ ∈ R,λ˜₂ = 0, that is a line passing through the origin.

Suppose thatκ₁ 6= 0andκ₂ 6= 0, andκ₁κ₂ >0, so they have the same sign. Then the only real solution to the quadratic equation (4.113) is the pointλ˜1 = ˜λ2 = 0, that is the origin.

63 4. Regularization of the inverse positioning problem

Suppose thatκ₁ 6= 0andκ₂ 6= 0, andκ₁κ₂<0, so they have different sign. Then the only real solution to the quadratic equation (4.113) is

˜λ₁ =±p

κ₂/κ₁˜λ₂, that is two lines passing through the origin.

Theorem 8. Suppose that for a spatial manipulator with the joint axes not meeting in one point either Q 6= 0 or dim KerJ_V^e = 2. Then J_V^e is regularizable if and only ifJ^eis full rank.

Proof. IfJ_V^e is regularizable, thenJ^eis full rank because of Theorem 6, so only the other direction needs to be proved. Suppose, thatJ^e is full rank. Then the regularized Jacobian is

J^reg=J_V^e +γ(J_Ω^e ×r), (4.114) which is regular if and only if

J^regλ= 0 (4.115)

implies λ = 0. It will be proved, that there exists r ∈ R³ and γ ∈ R, such that (4.115) holds if and only ifλ= 0. Note thatλ= 0 is a trivial solution, and throughout the proof, all nonzeroλs will be analyzed, and it will be shown, that ifλ6= 0, then for suitably chosenr ∈R³ andγ∈R (4.115) does not hold.

Substituting (4.114) into the condition (4.115), it can be rephrased asJ^reg is regular if and only if

J_V^eλ+γ(J_Ω^eλ)×r= 0 (4.116) impliesλ= 0. This condition can be further manipulated to get

J_V^eλ=−γ(J_Ω^eλ)×r. (4.117) Because of the properties of the vector product, this equation holds if and only if

hJ_V^eλ, ri = 0 (4.118)

hJ_V^eλ, J_Ω^eλi = 0 (4.119) and for allλ˜that is the solution of (4.118) and (4.119)

γ =−signD

J_V^eλ,˜ (J_Ω^eλ)˜ ×rE J_V^eλ˜

(J_Ω^eλ)˜ ×r

(4.120)

if(J_Ω^eλ)˜ ×r 6= 0.

64 4. Regularization of the inverse positioning problem

Let λ_V ∈ KerJ_V^e. In this caseJ_V^eλ_V = 0, so ensuring (J_Ω^eλ_V)×r 6= 0 guarantees that (4.117) does not hold. Since the joint axes of the ma-nipulator does not intersect at the same point,dim KerJ_V ≤2by Propo-sition 12. However, because of Lemma 2, J_Ω^eλ_V 6= 0 for any nonzero λ_V ∈ KerJ_V^e. Since KerJ_V^e is a subspace, and its dimension is not greater than two, the set {J_Ω^eλ_V : λ_V ∈ KerJ_V^e} is a plane, a line or a point containing the origin (in the last case it is the origin itself). If r is chosen such that it is perpendicular to {J_Ω^eλ_V :λ_V ∈ KerJ_V^e}, then (J_Ω^eλ_V)×r 6= 0, and (4.120) yields γ = 0 (by substituting λ˜ := λ_V).

So ifr is perpendicular to {J_Ω^eλ_V : λ_V ∈ KerJ_V^e}, then choosing γ 6= 0 guarantees that (4.117) does not hold for anyλ_V ∈KerJ_V^e.

Let S = (KerJ_V^e)^⊥be the orthogonal complement ofKerJ_V^e, i.e. S is a subspace ofR³,S∪KerJ_V^e =R³ and every vector fromSis orthogonal to every vector fromKerJ_V^e. Thus any vector fromR³ can be written as λ = c_Vλ_V +c_Sλ_S, with λ_V ∈ KerJ_V^e, λ_S ∈ S for some c_V, c_S ∈ R. For every nonzeroλS ∈S,J_V^eλS 6= 0, so(J_Ω^eλS)×r = 0ensures that (4.117) does not hold, so for everyλ_S ∈S, the condition(J_Ω^eλ_S)×r6= 0need not be satisfied. Note that even if (4.118) and (4.119) holds for someλ_S ∈S, but(J_Ω^eλS)×r= 0, this means thatJ_Ω^eλS andrare parallel, thus (4.117) does not hold.

Suppose that dim KerJ_V^e = 2. This yields that dim RanJ_V^e = 1and dimS= 1, thus the set{J_V^eλS :λS ∈S}is a one-dimensional subspace (a line passing through the origin) in the image space ofJ_V^e. Hovewerr is also in the image space ofJ_V^e because of Theorem 7, and since the image space is one-dimensional, hJ_V^eλS, ri 6= 0 for any nonzero λS ∈ S and nonzeror ∈R³, thus (4.118) can not hold for any nonzeroλ_S ∈S. Since J_V^e(c_Vλ_V +c_Sλ_S) =c_SJ_V^eλ_S, this concludes the proof ifdim KerJ_V^e = 2.

Ifdim KerJ_V^e <2, thenQ6= 0because of the condition of the theorem.

Notice, that because of the definition of matrix transpose, (4.119) can be rearranged to get D

λ,(J_V^e)^⊤J_Ω^e, λE

= 0. (4.121)

Because in a homogeneous quadratic form the matrix can be replaced with its symmetrical part, (4.121) can be written as

Dλ,1/2

(J_V^e)^⊤J_Ω^e +J_V^e(J_Ω^e)^⊤ , λE

= 0, (4.122)

and because of the definition ofQ, this equation simplifies to the quadratic form

hλ, Qλi= 0. (4.123)

Suppose, thatdim KerJ_V^e = 1. Then the condition (4.118) becomes hJ_V^e(c_Vλ_V +c_Sλ_S), ri=hJ_V^ec_Sλ_S, ri= 0. (4.124)

65 4. Regularization of the inverse positioning problem

Since dim KerJ_V^e = 1 implies dim RanJ_V^e = 2, and dimS = 2, the set {J_V^eλ_S : λ_S ∈ S} ⊆ RanJ_V^e is a two-dimensional subspace, while r ∈ RanJ_V^e spans a one-dimensional subspace (since it is a vector in the im-age space ofJ_V^e), so the setλ˜_Sfor which (4.124) holds is a one-dimensional subspace, lying inRanJ_V^e being orthogonal to the vector(J_V^e)^⊤r. Since

˜λ_Sis a one-dimensional subspace, it is a line passing through the origin.

Identifyλ˜_S with the direction vector of this line. Thus condition (4.123) needs to be examined only on the two-dimensional subspace spanned by the vectors˜λ_S and λ_V. Let this subspace be denoted byS. Then the˜ solution of the quadratic equation (4.123) restricted toS˜by Lemma 3 is either two lines, one line, or a point, all containing the origin. If there is a nontrivial solution (two lines or one line), then letλ˜ be the set of the unit direction vectors of the lines. Then ifγ is chosen such that (4.120) does not hold, then (4.117) does not hold, and this concludes the proof if dim KerJ_V^e = 1.

Suppose that dim KerJ_V^e = 0. In this casedimS = 3, and r can be chosen arbitrarily. Letr ∈R³be fixed. Then the condition (4.118) is only satisfied on the set{λ˜_S : ˜λ_S ∈S,˜λ_S⊥(J_V^e)^⊤r}, that is a two-dimensional subspace ofR³, denote it byS. Then the solution of (4.123) restricted to˜ S˜is either two lines, a line or a point containing the origin by Lemma 3. If there is a nontrivial solution (two lines or one line), then letλ˜ be the set of the unit direction vectors of the lines. Then ifγ is chosen such that (4.120) does not hold, then (4.117) does not hold, and this concludes the proof ifdim KerJ_V^e = 0.

4.3.4 The determinant and the singular values of the

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