上海交通大学：《Design and Manufacturing》课程教学资源（讲义）Lecture 6 Vector fundamentals for kinematic analysis

园 上活我大峰 eb口 LECTURE 6 Vector fundamentals for kinematic analysis 1日 国 上泽充通大警 OUTLINE Why kinematic analysis Space of motion Vector manipulations Motion as vectors Instantaneous velocity center ME337 Design Manufacturing ll

Vector fundamentals for kinematic analysis LECTURE 6 ME337 Design & Manufacturing II OUTLINE Why kinematic analysis Space of motion Vector manipulations Motion as vectors Instantaneous velocity center

园 上活我大峰 Why kinematic analysis x Force x Force x Velocity x Velocity ×Torque Input Output x Torque P Power Energy Link 2 angle 54.00 1.64 3.30 0.26 X-Position of Link41可 Velocity of Link41▣ Acceleration of Link 4 11 .×m 10 Vx [enisl -0 A m) 5.0 2 2.0 5.D 2.0 10 5.016 15 20 国 上游文通大学 Why kinematic analysis To analyze the displacement,velocity,and acceleration of the motion to determine the design geometry of the mechanical parts. To determine the motion of a rigid body caused by the forces applied to it

Why kinematic analysis Why kinematic analysis To analyze the displacement, velocity, and acceleration of the motion to determine the design geometry of the mechanical parts. To determine the motion of a rigid body caused by the forces applied to it

国上活大峰 © Displacement analysis 1)to determine the output displacement and check if it meets the requirements. 2)to find the stroke or output workspace. 3)to avoid space interferences. ©Velocity analysis 1)the basis for further analysis 2)to find the velocity characteristics of the output,then check if it meets the requirements. (such as quick return mechanism) 圆上泽大华 Acceleration analysis 1)to find the inertial forces 2)to determine the dynamic performances of the output

Displacement analysis 1) to determine the output displacement and check if it meets the requirements. 2) to find the stroke or output workspace. 3) to avoid space interferences. Velocity analysis 1) the basis for further analysis 2) to find the velocity characteristics of the output, then check if it meets the requirements. (such as quick return mechanism…) Acceleration analysis 1) to find the inertial forces 2) to determine the dynamic performances of the output

图上泽通大学 ANALYSS OF THE FOURAR LNKAGE 0+ 锦 国上活庆大峰 Kinematic analysis methods IC ●Graphical Graphical method based on vectors Complex vector method ●Analytical Matrix

Kinematic analysis methods ●Graphical ●Analytical IC Graphical method based on vectors Complex vector method Matrix

国 上浒充通大￥ Space of motion Space of motion 国上序庆大￥ Space of motion Curvilinear translation General plane motion o Rectilinear translation Rotation about a fixed axis fig16_02.jpg Copyright 2010 Pearson Prentice Hall,Inc

Space of motion Space of motion Space of motion

园 上活我人峰 Type of Rigid-Body Plane Motion Example a Rectilinear translation Rocket test sled 6) Parallel-link swinging plate (e) Compound pendulum d Seaan Connecting rod in a reciprocating engine 圈上清庆大坐 Scalars Vectors Examples: mass,volume,speed force,velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font,a line,an arrow or a“carrot

Scalars Vectors Examples: mass, volume, speed force, velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font, a line, an arrow or a “carrot

国上活大峰 Vector manipulations Vector Magnitude Head Magnitude (terminus) Direction Tail (origin) Vector can be represented Graphically rR=Ri+Ryj R=R,+jRy Analytically R=[RR,T R=Reio=R(cos0+jsin) 圈上清庆大坐 Vector manipulations Vector addition: A=Aeio B=Bei0s C=A+B C=A+B=Aeio+Beia =B+A =(Acos0+Bcoseg)+ j(Asin0+Bsin0g) AC-B Vector substraction: A=C-B=Cei@c-Beia =(Ccos0-Bcose)+ j(Csinc-Bsineg) Figure (a)Vector addition and (vector subtraction

Vector manipulations Vector  Magnitude  Direction Vector can be represented  Graphically  Analytically Head Magnitude (terminus) Tail (origin) Rx y R   jR R e (cos sin ) j R j  R     [ ]T R  R x y R Rx y R  i R  j Vector manipulations Figure (a) Vector addition and (b) vector subtraction.  B  A  B  C A j Ae  A  B j Be  B  =( cos cos )+ j( sin sin ) A B j j A B A B Ae Be A B A B           CAB =( cos cos )+ j( sin sin ) C B j j C B C B Ce Be C B C B           A CB Vector addition: Vector substraction:

国上活大峰 0 B (a) Figure Graphical solution of case 2:(a)given C.A.and B. Figure Graphical solution of case 3 (a)given C.A'.and B': c的solution for A,B°andA'.B". 句solution for A and B. √√o√√o C=A+B 6=A+B 圈上清庆大坐 Vector rotation: R=Reio R R'=Reio=Rei(0+o) R

Figure Graphical solution of case 2: (a) given C, Aˆ , and B; (b) solution for A, Bˆ and A’, Bˆ’. CAB   √ √ O √ √ O √ √ O √ O √ Figure Graphical solution of case 3: (a) given C, Aˆ, and Bˆ; (b) solution for A and B. CAB   Vector rotation: j R e  R  ( ) 'e j j R e    R R  X Y O R  R ' 

国上清大峰 o=i2+j(-1)+k4(rad1s). Vector cross product: r=i(-1)+j10+k2(mm), A=iAs jA,+kA. B=iB +jBy+kB. =i(-2-40)+j-4-4)+k(20-1) i jk =i(-42)+j-8)+k19)mm/s A×B=AA,A BB B. AxB=i(A,B.-A.B)+j(A.B,-AB.)+k(A,B-A,B, A×B=ABsin0 圆上清夫大学 Vector dot product: A[B=ABcos0 Angle 0 is the smallest angle between the two vectors and is always in a range of0°to180°

ω  i2  j(1)  k4 (rad / s). r  i(1)  j10  k2 (mm), mm s B B B A A A x y z x y z ( 42) ( 8) (19) / ( 2 40) ( 4 4) (20 1) 1 10 2 2 1 4 i j k i j k i j k i j k υ ω r                    ω r v Ax Ay A z A  i  j  k Bx By B z B  i  j  k Vector cross product: x y z x y z B B B A A A i j k AB  ( ) ( ) ( )   AyBz  AzBy  AzBx  AxBz  AxBy  AyBx A B i j k A B  ABsin Vector dot product: A B￾  ABcos Angle  is the smallest angle between the two vectors and is always in a range of 0º to 180º

国上活天大峰 -1- ixj=k jxi=-kj×k=ikxi=-i kxi=j ixk=-j i×i=j×j=k×k=0 i.i=jj=k.k=1 i.j=j.k=k.i=0 圆上泽夫大峰 commutative BxA=-AxB. A.B=B.A Distributive AX(B+C)=A×B+A×C A.(B+C)=A.B+A.C Triple Product 44,4 A(B×C)=B.(C×A)=C.(A×B)=BB,B c.c,c. A×(B·C)=(A.C)B-(A.B)C (A×B)×C=(A.C)B-(B.C)A

           0           k i j i k j i i j j k k i j k j i k j k i k j i i.i  j.j  k.k 1 i.j  j.k  k.i  0 BA  AB. A.B  B.A A (B  C)  A B  AC A.(B  C)  A.B  A.C commutative Distributive ( )( )( ) ; xyz xyz x y z A A A B B B CCC A. B C B. C A C. A B    ( )( ) ( ) ( ) ( )()      A B C A.C B A.B C A B C A.C B B.C A Triple Product