Three Problems in Robotics - helix

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this could, of course, also have been deduced from elementary mechanics.For the spring systems of section 2 an important object is the stiffness matrix of thesystem. In this section we show how to compute this by taking the second derivativesof the potential function.An infinitesimal displacement of the body is represented by a screw. The wrenchproduced by a displacement s is given by W = Ks, where K is the stiffness matrix.The stiffness matrix is the hessian of the potential function, that is its matrix ofsecond order partial derivatives, see [1, Ch 5]. This is only valid at an equilibriumposition.There have been attempts in the Robotics literature to extend these ideas to nonequilibriumconfigurations, see for example Griffis and Duffy [4] and Žefran and Kumar[18]. In this work, however, the classical definition of the stiffness matrix will beused.Now let us write the result above for the wrench as( )Ai 0 )W = ∑λ iIi 3 0(ãi − M ˜b iDifferentiating this along an arbitrary screw gives,( )(Ai 0 RBi R∂ S W = ∑λ T )+ T I 3 ωiIi 3 0 0 0)(vusing the result of section 2. Hence, we have that the stiffness matrix is,( )∑i λ i (A i RB i R T + A i T ) ∑ i λ i A iK =∑ i λ i (RB i R T + T ) ∑ i λ i I 3This time we choose the origin to be at the point ∑ i λ i a i and then use the equilibriumcondition 2 to simplify the stiffness matrix to,( )∑i λ i A i RB i R T 0K =0 ∑ i λ i I 3This is a particularly neat result but it is a little surprising at first sight. The term∑ i λ i I 3 in the bottom right corner means that the system has the same stiffness in anydirection, irrespective of the stiffness of the individual springs and their arrangement.This is due to the assumption that the springs have zero natural length.Now we can return to the problem of finding the index of the critical points we havefound. For brevity we will just look at which of the critical points is a minimum of thepotential energy, that is a stable equilibrium. That is: which of the solutions for R givesa stiffness matrix K with all positive eigenvalues? Notice that three of the eigenvaluesof K are simply ∑ i λ i , and this is positive if the spring constants λ i are all positive. Sowe just have to look at the eigenvalues of the top-lefthand bock of K. After a littlemanipulation this can be written in term of the matrix P or its polar decomposition,∑λ i A i RB i R T = RP T + Tr(RP T )I 3 = R i Q + Tr(R i Q)I 3iHere Tr() represents the trace of the matrix. Now this matrix has the same eigenvectorsas R i Q and hence as Q, remember R i was either the identity or a rotation of π about9
an eigenvector of Q. So let us assume that the eigenvalues of Q are µ 1 , µ 2 and µ 3with corresponding eigenvectors e 1 , e 2 and e 3 . Assuming that R i is a rotation abouteigenvector e i , we can find the eigenvalues of the matrices by considering the action onthe eigenvectors e 1 , e 2 and e 3 . The eigenvalues for R 0 Q + Tr(R 0 Q)I 3 are,(2µ 1 + µ 2 + µ 3 ), (µ 1 + 2µ 2 + µ 3 ), and (µ 1 + µ 2 + 2µ 3 )We only need to look at one other matrix since the other are just cyclic permutations,the eigenvalues of R 1 Q + Tr(R 1 Q)I 3 are,(2µ 1 − µ 2 − µ 3 ), (µ 1 − 2µ 2 − µ 3 ), and (µ 1 − µ 2 − 2µ 3 )Now there are just two cases to consider, if det(P) > 0 then we can assume that Qis positive-definite and thus so are all its eigenvalues. In this case it is easy to see thatthe critical point represented by R 0 will be the minimum. This is the solution R = R p .In the other case det(P) < 0, the classical polar decomposition gives us a positivedefinitesymmetric matrix and a reflection. When we multiply these by −1 wegetarotation and a negative-definite symmetric matrix. That is Q has all negative eigenvalues.If we assume that these eigenvalues have the ordering 0 > µ 1 ≥ µ 2 ≥ µ 3 , that is µ 1is the eigenvalue of smallest magnitude, then the matrix, R 1 Q + Tr(R 1 Q)I 3 will haveall positive eigenvalues and hence R 1 corresponds to the minimum of the potential. Soin general, if det(P) < 0 the minimum solution is given by R = R p R i , where R i is arotation of π about the eigenvector of Q with eigenvalue of smallest magnitude.6 ConclusionsThe concept unifying the three problems we have studied in this work is the idea offunctions defined on the group of rigid body transformations. We have been able to findthe stationary points of the functions and classify these critical points. This involves asimple technique of differentiating along a screw.The results agree with those of Kanatani [7, Chap. 5], his methods were, perhaps,a little more elegant than the above. But the methods used here are more general, wehave found all the critical points of the potential function not just the minimum andwith a little more effort we could find the index of each of them.These ideas are central to the subject of Morse theory, a field of study which relatesthe critical points of functions defined on a manifold to the topology of the manifolditself. In this case the topology of the manifold, the underlying manifold of the groupSO(3), is well known, and we can use this to say something about the critical point ofour potential function.The problem of the springs is a slightly artificial one in that we have used springswill zero natural length. This simplifies the computations. It is possible to repeat mostof the above analysis using spring with a finite natural length, see [14]. The numberof critical points now becomes a very hard problem but will still be constrained by thetopology of the group.The problem of estimating a rigid motion from point data is reasonably realistic.The utility of the estimate will depend on the distribution of error for the measuredpoints. There are some results on this in the literature, see [17]. However, I believethere is much scope for further research in this direction.10
Page 1 and 2: Three Problems in RoboticsJ.M. Seli
Page 3 and 4: uration. If the body undergoes a ri
Page 5 and 6: into an orthogonal matrix and a non
Page 7 and 8: onal and D diagonal. Now we can wri
Page 9: So we see that the rows of this Jac
Page 13: [15] J.M. Selig and P.R McAree. Con

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