LoesungA5_2001 - 4 ( cos γ sin α + sin γ cos α | {z } =...

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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 A5 Seite 1 Musterl¨ osung Aufgabe 5 ω 1 ω 3 1 ω 2 3 ω 2 2 ω 4 0 ω 4 x y 1 2 3 4 α β β γ n 1 Gegeben : n 1 ,α,β Zusammenhang Drehzahl - Winkelge- schwindigkeit ω = 2 π T = 2 πν = 2 π n 60 a) Gesucht : n 3 Schliessbedingungen ader 1-2-4: 0 ω 1 + 1 ω 2 + 2 ω 4 = 0 ω 4 ! = 0 (1) ader 3-2-4: 0 ω 3 + 3 ω 2 + 2 ω 4 = 0 ω 4 ! = 0 (2) Komponentendarstellung beider Gleichungen in x- und y-Richtung (1) in x: - ω 1 + 1 ω 2 cos α - 2 ω 4 cos γ = 0 (3) (1) in y: - 1 ω 2 sin α - 2 ω 4 sin γ = 0 (4) (2) in x: - ω 3 + 3 ω 2 cos( β + γ ) - 2 ω 4 cos γ = 0 (5) (2) in y: - 3 ω 2 sin( β + γ ) - 2 ω 4 sin γ = 0 (6) = 4 Gleichungen mit 4 Unbekannten 1 ω 2 , 2 ω 4 , 3 ω 2 3 (3) · sin α + (4) · cos α : - ω 1 sin α - 2 ω
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Unformatted text preview: 4 ( cos γ sin α + sin γ cos α | {z } = sin( α + γ ) ) = 0 = ⇒ 2 ω 4 =-sin α sin( α + γ ) ω 1 (7) (5) · sin( β + γ ) + (6) · cos( β + γ ) :-ω 3 sin( β + γ ) (8)-2 ω 4 ( cos γ sin( β + γ )-sin γ cos( β + γ ) | {z } = sin( β + γ-γ ) = sin β ) = 0 = ⇒ ω 3 =-sin β sin( β + γ ) 2 ω 4 (9) = ⇒ n 3 =-sin α sin β sin( α + γ ) sin( β + γ ) n 1 (10) b) Gesucht : 3 ω 2 aus (6): 3 ω 2 =-sin γ sin( β + γ ) 2 ω 4 = ⇒ 3 ω 2 =-sin α sin γ sin( α + γ ) sin( β + γ ) ω 1 (11)...
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This note was uploaded on 03/15/2012 for the course MAVT KIN07 taught by Professor Hiller during the Spring '01 term at Swiss Federal Institute of Technology Zurich.

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