ࡱ> ;=: @ l]bjbj c'<<<<4HHHhHI2JK(LLBL'.Ui $RL=Eu6 'uu=<<LBLfkkku<( L:BLkukkHd gBLJ `LٺHc40/8q(g <<<<g|uukuuuuu==6HI"HPosition Analysis: Review (Chapter 2)  Objective: Given the geometry of a mechanism and the input motion, find the output motion Graphical approach Algebraic position analysis Example of graphical analysis of linkages, four bar linkage. Given a-d and (2 find (3 and (4       SHAPE \* MERGEFORMAT   SHAPE \* MERGEFORMAT   SHAPE \* MERGEFORMAT  Algebraic position analysis Use trigonometry to find positions of links and joints. Example: four bar linkage:      Procedure: Find triangle O2AO4 from a, d, (2. Find triangle AO4B from b, c, AO4. Complex number method for position analysis Idea: represent links as position vectors and represent these vectors as complex numbers. Why complex numbers?  In many problems, it is more straightforward to derive the equations for position analysis using complex numbers Complex number: Rectangular form     Complex number: Polar form     Complex conjugate EMBED Equation.3 Eulers theorem: EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 EMBED Equation.3 Complex number algebra: EMBED Equation.3 Equality: z = w EMBED Equation.3 Addition, subtraction: EMBED Equation.3 Multiplication, Division, Powers Multiplication of two numbers: multiply magnitudes, add phase angles. Division of two numbers: divide magnitudes, subtract phase angles Raising a complex number into a power: raise magnitude into the power, multiply phase angle by the power Use rectangular form when adding or subtracting Use polar from when multiplying, dividing or raising into a power Complex equation solving: f(z) = 0, where z and f are complex quantities. Solve for z. Solution Real(f(z)) = 0 or Real[f(x+jy)] = 0 Im(f(z)) = 0 or Im[f(x+jy)] = 0 Solving the above two independent equations for x and y we find z. Example: (4+j)z+5-2j = 0, where z = x + jy Solution Substituting z = x + jy into the equation we obtain: (4+j)(x+jy)+5-2j = 0 or 4x+4y j +xj-y+5-2j=0 or 4x-y+5+j(4y+x-2)=0 Both real and imaginary parts should be zero: Real part = 0 ! 4x - y+5=0 Imaginary part = 0 ! 4y+x-2=0 Solving the two equations for x and y we obtain the real and imaginary parts of complex number z: x = -1.059 and y = 0.765. Therefore: z = -1.059 + j0.765. Example: crank-rocker four bar linkage  Problem: Given a, b, c, d and (2, find (3 and (4 Solution: Vector loop equation:  EMBED Equation.3  Real part = 0 Imaginary part = 0 Two equations with two unknowns, (3 and (4 Important definitions: Open Grashof mechanism: If  EMBED Equation.3 then the two links adjacent to the shortest link (crank) do not cross each other. Crossed Grashof mechanism: If  EMBED Equation.3 then the two links adjacent to the shortest link (crank) cross each other.  EMBED Equation.3  EMBED Equation.3  Open solution for negative square root, crossed solution for positive square root Open solution for negative square root, crossed solution for positive square root Example: Slider-crank mechanism Problem: Given a-d, (2 find b, (3, (4 Open solution first  Steps:  EMBED Equation.3  Final result: Find angle  EMBED Equation.3  by solving numerically or algebraically the following equation:   EMBED Equation.3  Algebraic solution: Open solution  EMBED Equation.3  Inclination angle of coupler:  EMBED Equation.3  End of open solution Crossed mechanism:  EMBED Equation.3    EMBED Equation.3  Algebraic solution:  EMBED Equation.3  Find coefficients T, S and U from the equations for the open solution. 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