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LoesungA7de_2001

# LoesungA7de_2001 - Gerhard-Mercator-Universitt Duisburg...

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Gerhard-Mercator-Universitt Duisburg Fachgebiet Mechatronik Prof. Dr.-Ing. habil. M. Hiller Computergest¨utzte Kinematik und Dynamik von Mechanismen WS 2001/02 osung A7 Seite 1 Musterl¨ osung Aufgabe 7 d) Gesucht : Euler-Parameter q 0 , q 1 , q 2 und q 3 (Vorgehen wie bei Drehzeiger) T Q = q 2 0 I + 2 q q - q · q I | {z } + 2 q 0 q × | {z } symmetrischerAnteil schiefsym . Anteil T s Q = 1 2 T Q + T T Q · T a Q = 1 2 T Q - T T Q · Skript (4.89) 1) Element q 0 aus der Spur von T Q t 11 + t 22 + t 33 = ( q 2 0 - q · q ) (1 + 1 + 1) + 2 ( q 2 1 + q 2 2 + q 2 3 ) = 3 q 2 0 - 3 ( q 2 1 + q 2 2 + q 2 3 ) + 2 ( q 2 1 + q 2 2 + q 2 3 ) = 3 q 2 0 - q · q N ( Q ) = 1 = q 2 0 + q · q q · q = 1 - q 2 0 Skript (5.58) t 11 + t 22 + t 33 = 4 q 2 0 - 1 q 0 = ± 1 2 1 + t 11 + t 22 + t 33 = cos β 2 (Vorzeichen w¨ ahlbar) (1) 2) Elemente q 1 , q 2 , q 3 aus schiefsymmetrischem Anteil T a Q = 2 q 0 0 - q 3 q 2 q 3 0 - q 1 - q 2 q 1 0 | {z } = 1 2 T Q - T T Q · = 1 2 0 t 12 - t 21 t 13 - t 31 t 21 - t 12 0 t 23 - t 32 t 31 - t 13 t 32 - t 23 0 | {z } Ein Komponentenvergleich ergibt: q 1 = 1 4 q 0 ( t 32 - t 23 ) q 2 = 1 4 q 0 ( t 13 - t 31 ) (2) q 3 = 1 4 q 0 ( t 21 - t 12 ) mit q 0 = ± 1 2 1 + t 11 + t 22 + t

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