Lecture note provided by Dr. Daejong Kim, MAE 3318 Ch. 7 Acceleration analysis Pure rotation RP ae j2 --> VP aj2e j2 --> AP aj2e j2 a2 2e j2 Rotation and translation RP RA ae j2 --> VP VA aj2e j2 --> AP AA aj2 a2 2 e j2 VP / A AP / A AP AA AP / A : total acceleration of point P is a combination of acceleration of point A and relative acceleration of P with respect to A. Four bar linkage Procedure Identify inputs and outputs - Input: position, velocity, and acceleration of link 2 - Output (unknowns): accelerations of link 3 and 4 or any point within the system ae j2 be j3 ce j3 de j1 0 Start with vector loop equation: Differentiate with time twice (3 and 4 are variables) aj a e bj b e cj c e 2 2 j 2 2 AA 2 3 3 AB / A j3 2 4 j 4 4 AB AA AB / A AB The first two terms on the left hand side are nothing but total acceleration of point B, and the term on the right hand side is total acceleration of point B when you evaluate from link 4 (just pure rotation wrt O4). Above equation just tells us equality of total acceleration of point B whether you find it by adding AA and AB/A or directly evaluating from link 4. Insert Euler form aj2 (cos 2 j sin 2 ) a2 2 (cos 2 j sin 2 ) bj (cos j sin ) b 2 (cos j sin ) 3 3 3 3 3 3 cj4 (cos 4 j sin 4 ) c4 (cos 4 j sin 4 ) 0 2 Separate real and imaginary parts a2 sin 2 a2 2 cos 2 b3 sin 3 b32 cos3 c4 sin 4 c4 2 cos 4 0 a cos a 2 sin b cos b 2 sin c cos c 2 sin 0 2 2 2 2 3 3 3 3 4 4 4 4 Rearranging above, b3 sin 3 c4 sin 4 a2 sin 2 a2 2 cos 2 b32 cos3 c4 2 cos 4 b3 cos3 c4 cos 4 a2 cos 2 a2 2 sin 2 b32 sin 3 c4 2 sin 4 : This is 2x2 algebraic equations for two unknowns Putting into matrix form, b sin 3 b cos3 c sin 4 3 a2 sin 2 a2 2 cos 2 b32 cos3 c4 2 cos 4 c cos 4 4 a2 cos 2 a2 2 sin 2 b32 sin 3 c4 2 sin 4 Solving for 3 3 , 4 4 A c sin 4 --> CD AF 3 AE BD where CE BF 4 AE BD B b sin 3 C a 2 sin 2 a2 2 cos 2 b32 cos3 c4 2 cos 4 D c cos 4 E b cos3 F a 2 cos 2 a2 2 sin 2 b32 sin 3 c4 sin 4 Four bar slider-crank Procedure Identify inputs and outputs - Input: position, velocity, and acceleration of link 2 - Output (unknowns): accelerations of link 3 and 4 (slider) or any point within the system ae j2 be j3 ce j3 de j1 0 Start with vector loop equation Differentiate with time twice (d and 3 are variables) aj a e 2 2 2 j 2 bj3 b32 e j3 d 0 Inserting Euler form aj2 (cos 2 j sin 2 ) a2 2 (cos 2 j sin 2 ) bj3 (cos 3 j sin 3 ) b32 (cos 3 j sin 3 ) d 0 Separate real and imaginary parts a2 sin 2 a2 2 cos 2 b3 sin 3 b32 sin 3 d 0 a cos a 2 sin b cos b 2 sin 0 2 2 2 2 3 3 3 : This is 2x2 algebraic equations for 3 two unknowns --> a 2 cos 2 a2 2 sin 2 b32 sin 3 3 b cos3 d a 2 sin 2 a2 2 cos 2 b 3 sin 3 b32 cos 3 Coriolis acceleration Example where Corilois acceleration is negative direction ( V p _ slip V p _ radial , Ap _ slip Ap _ radial ) AP 2 AP Coriolis APn VP AP AP Coriolis APslip APt Example where Corilois acceleration is positive direction ( V p _ slip V p _ radial , Ap _ slip Ap _ radial ) RP pe j2 Define R p in terms of radius and angle 2 Take derivative of R p with respect to time Take derivative of R p with respect to time twice j2 p2 je j2 V p _ tran V p _ slip VP pe AP pe j2 p2 2e j2 2 p2 je j2 p2 je j2 Ap Ap _ slip Ap _ normal Ap _ Coriolis Ap _ tan gential Physically interpret each terms in vector diagram <-- Ap _ slip Ap _ radial Inverted slider-crank: 3 A a B b c 2 4 Text book notation Constraint equation: 3 4 . We will use the following vector directions and new angle to describe the motion. Note the direction of vector for link 3 is now from A to B. This configuration is easier to visualize the Coriolis acceleration component of link 4 with respect to A. A a b B c 2 Prefered vector and angle notation of link 3 4 Velocity vector loop equation j cj e j4 0 aj2 e j2 bj e j be 4 Acceleration vector loop equation a 2 je j2 a2 2 e j2 b b 2 e j b 2b je j c 4 je j4 c4 2 e j4 0 a j a e (b b ) (b 2b ) j e c j c e j 2 2 2 j 2 2 2 4 AA j 4 4 AB AB / A The first two terms on the left hand side are nothing but total acceleration of point B, and the term on the right hand side is total acceleration of point B when you evaluate from link 4 (just pure rotation wrt O4). Above equation just tells us equality of total acceleration of point B whether you find it by adding AA and AB/A or directly evaluating from link 4. Inserting Euler form, a j a (cos j sin ) (b b c j c (cos j sin ) 2 2 2 2 2 2 ) (b 2b ) j (cos j sin ) 2 4 4 4 4 Separating real and imaginary parts from above equations, you can find expression for b and ( ) when 2 , 2 are inputs. 4 7.7 Acceleration of arbitrary points in the linkage Once you have found 3 4, acceleration of any point within the linkage can be found by evaluating position vectors and taking time derivative twice. Position S: RS se j (2 2 ) Position P: RP ae j2 pe j (3 3 ) Position U: RU ue j (4 4 ) Example: Acceleration of point P RP ae j2 pe j (3 3 ) AP a 2 je j2 a2 2 e j2 p 3 je j (3 3 ) p32 e j (3 3 ) a 2 j (cos 2 j sin 2 ) a2 2 (cos 2 j sin 2 ) p 3 j (cos(3 3 ) j sin(3 3 )) p32 (cos(3 3 ) j sin(3 3 )) a 2 sin 2 a2 2 cos 2 p 3 sin(3 3 ) p32 cos(3 3 ) j a 2 cos 2 a2 2 sin 2 p 3 cos(3 3 ) p32 sin(3 3 ) APX a 2 sin 2 a2 2 cos 2 p 3 sin(3 3 ) p32 cos(3 3 ) APY a 2 cos 2 a2 2 sin 2 p 3 cos(3 3 ) p32 sin(3 3 )