Example: EQS5

Four-bar linkage kinematic analysisFor and `φ`_{3}(0) = 0.789 [rad], `φ`_{4}(0) = 4.492 [rad] solve the algebraic equations characterizing kinematics of the four-bar mechanism:

0 | = | r_{2} cos (φ_{2}) + r_{3} cos (φ_{3}) – r_{4} cos (φ_{4}) – r_{1} |

0 | = | r_{2} sin (φ_{2}) + r_{3} sin (φ_{3}) + r_{4} sin (φ_{4}) |

Position of the critical point T in the (x, y)-plane can be computed by evaluating the expressions:

x_{T} | = | r_{2} cos (φ_{2}) + r_{3}/2 cos (φ_{3}) |

y_{T} | = | r_{2} sin (φ_{2}) + r_{3}/2 sin (φ_{3}) |

The mechanism is driven by the constant angular velocity of bar 2

r1 | 0.6 | r_{1} = 0.6 | [m] | distance of hinges |

r2 | 0.3 | r_{2} = 0.3 | [m] | length of bar 2 |

r3 | 0.55 | r_{3} = 0.55 | [m] | length of bar 3 |

r4 | 0.4 | r_{4} = 0.4 | [m] | length of bar 4 |

March 14, 2008