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By Leila Notash (auth.), Federico Thomas, Alba Perez Gracia (eds.)

ISBN-10: 9400772130

ISBN-13: 9789400772137

ISBN-10: 9400772149

ISBN-13: 9789400772144

Computational kinematics is a charming quarter of technology with a wealthy spectrum of difficulties on the junction of mechanics, robotics, laptop technological know-how, arithmetic, and special effects. The coated issues comprise layout and optimization of cable-driven robots, research of parallel manipulators, movement making plans, numerical tools for mechanism calibration and optimization, geometric ways to mechanism research and layout, synthesis of mechanisms, kinematical concerns in biomechanics, development of novel mechanical units, in addition to detection and remedy of singularities.

The effects could be of curiosity for training and study engineers in addition to Ph.D. scholars from the fields of mechanical and electric engineering, machine technology, and machine graphics.

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3 Related Historical Work Before we list related historical results, we repeat some elementary facts on regular ruled quadrics, which are the one-sheeted hyperboloid and the hyperbolic paraboloid (see Fig. 1): Both surfaces carry two sets of generators, which are called regulus R and associated regulus R× , respectively. Moreover, it should be noted that all lines within one set are skew to each other and that each line of one set is intersected by 52 G. Nawratil all lines of the other set. Therefore, a regular ruled quadric is uniquely determined by three pairwise skew generators.

1 The points N4 , N5 , N6 are pairwise distinct and do not belong to s = sκ. Moreover, N4 , N5 , N6 are not collinear. Proof The point Nk+3 is located on the line [nk+3 , Nk+3 ], which belongs to the reguli Ri and R j for pairwise distinct i, j, k ∈ {1, 2, 3}. Therefore, these three lines [n4 , N4 ], [n5 , N5 ], [n6 , N6 ] are pairwise skew and not located within the platform πm . As a consequence, the points N4 , N5 , N6 are pairwise distinct and not located on s = sκ. Now, we prove the second part of this lemma by contradiction.

Fig. 3b): We consider any finite point S ∈ s = sκ. This point spans together with the ideal points f ∈ πm and F ∈ πM (cf. 2 Now we intersect α with a plane β, which contains S and is orthogonal to the direction f. We denote the line of intersection by t. Then we chose ε as the plane spanned by s = sκ and t. 2, the resulting planar projective SG platform with platform anchor points n1 , . . , n6 and base anchor points N1 , . . , N6 possesses an elliptic 2 Note, that f ∈ s or F ∈ sκ cannot hold as this yields f = F, a contradiction.

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Computational Kinematics: Proceedings of the 6th International Workshop on Computational Kinematics (CK2013) by Leila Notash (auth.), Federico Thomas, Alba Perez Gracia (eds.)


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