Notes_03_02

# Notes_03_02 - Notes_03_02 1 of 15 Complex Numbers for...

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Notes_03_02 1 of 15 Complex Numbers for Planar Kinematics Standard XY Notation Complex Number Notation θ = + = r j ˆ b i ˆ a R θ = + = j e r b j a R 2 2 2 b a r + = θ + θ = θ sin j cos j e a / b tan = θ 1 j - = Expanding: θ + θ = θ = sin r j cos r j re R θ = cos r a θ = sin r b direction j e , magnitude r = θ = Position: θ = j re R Velocity: θ θ + θ = j e r j j e r R ϖ = θ = θ dt d α = θ = θ 2 2 dt d Extension Tangential Acceleration: θ θ + θ θ - θ θ + θ = j e r 2 j j e r j e r j j e r R 2 Extension Tangential Normal Coriolis Jerk: θ θ + θ + θ - θ + θ θ - θ θ - = j e j ) r 3 r 3 r r ( j e ) r 3 r 3 r ( R 3 2 Note: 1. θ is always measured CCW from positive real axis. θ θ θ and , are positive CCW. 2. θ θ j e j e j (Try it using θ = + = j e r b j a R ) Matrix Solution: [A]{x} = {B} {x}=[A] -1 {B} 21 12 22 11 11 21 12 22 1 22 21 12 11 a a a a ] A det[ a a a a ] A det[ 1 ] A [ a a a a ] A [ - = - - = = - X Y Real (Horizontal) Imaginary (Vertical) b a r θ b a r θ

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Notes_03_02 2 of 15 Complex Number Analysis of Four Bar Given: constants r 1 r 2 r 3 r 4 θ 1 and variable θ 2 Find: θ 3 and θ 4 Loop Equation: 0 R R R R 1 4 3 2 = - - + Position: 0 j e r j e r j e r j e r 1 1 4 4 3 3 2 2 = θ - θ - θ + θ Real Components: 0 cos r cos r cos r cos r 1 1 4 4 3 3 2 2 = θ - θ - θ + θ Imaginary Components: 0 sin r j sin r j sin r j sin r j 1 1 4 4 3 3 2 2 = θ - θ - θ + θ Use Newton-Raphson iterative solution for position. Given: Position solution and 0 r r r r and 1 4 3 2 1 2 = θ = = = = θ Find: 3 3 and θ θ Velocity: 0 j e r j j e r j j e r j 4 4 4 3 3 3 2 2 2 = θ θ - θ θ + θ θ Real Components: 0 sin r sin r sin r 4 4 4 3 3 3 2 2 2 = θ θ + θ θ - θ θ - Imaginary Components: 0 cos r j cos r j cos r j 4 4 4 3 3 3 2 2 2 = θ θ - θ θ + θ θ Matrix Notation: θ θ - θ θ = θ θ θ - θ θ θ - 2 2 2 2 2 2 4 3 4 4 3 3 4 4 3 3 cos r sin r cos r cos r sin r sin r [ ] ( 29 ( 29 γ - = θ - θ = θ θ - θ θ = sin r r sin r r sin cos cos sin r r J det 4 3 4 3 4 3 4 3 4 3 4 3 ( 29 ( 29 4 3 3 4 2 2 2 3 sin r / sin r θ - θ θ - θ θ - = θ ( 29 ( 29 4 3 4 3 2 2 2 4 sin r / sin r θ - θ θ - θ θ - = θ Given: Position and velocity solutions and 2 θ Find: 3 θ and 4 θ Acceleration: 0 j e r j e r j j e r j e r j j e r j e r j 4 2 4 4 4 4 4 3 2 3 3 3 3 3 2 2 2 2 2 2 2 = θ θ + θ θ - θ θ - θ θ + θ θ - θ θ Real: 0 cos r sin r cos r sin r cos r sin r 4 2 4 4 4 4 4 3 2 3 3 3 3 3 2 2 2 2 2 2 2 = θ θ + θ θ + θ θ - θ θ - θ θ - θ θ - Imaginary: 0 sin r j cos r j sin r j cos r j sin r j cos r j 4 2 4 4 4 4 4 3 2 3 3 3 3 3 2 2 2 2 2 2 2 = θ θ + θ θ - θ θ - θ θ + θ θ - θ θ Matrix: θ θ - θ θ + θ θ + θ θ - θ θ - θ θ + θ θ + θ θ = θ θ θ - θ θ θ - 4
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