# KND Ch7 updated rev3

```Lecture note provided by Dr. Daejong Kim, MAE 3318
Ch. 7 Acceleration analysis
Pure rotation
RP  ae j2 --&gt; VP  aj2e j2
--&gt; AP  aj2e j2  a2 2e j2
Rotation and translation


RP  RA  ae j2 --&gt; VP  VA  aj2e j2 --&gt; AP  AA  aj2  a2 2 e j2




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.
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 j2  be j3  ce j3  de j1  0


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
j3
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

Insert Euler form
aj2 (cos 2  j sin  2 )  a2 2 (cos 2  j sin  2 )
bj (cos  j sin  )  b 2 (cos  j sin  )
3
3
3
3
3
3
cj4 (cos 4  j sin  4 )  c4 (cos 4  j sin  4 )  0
2

Separate real and imaginary parts
 a2 sin  2  a2 2 cos 2  b3 sin 3  b32 cos3  c4 sin  4  c4 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,
b3 sin 3  c4 sin  4  a2 sin  2  a2 2 cos  2  b32 cos3  c4 2 cos 4
b3 cos3  c4 cos 4  a2 cos 2  a2 2 sin  2  b32 sin 3  c4 2 sin  4
: This is 2x2 algebraic equations for two unknowns
Putting into matrix form,
 b sin 3

 b cos3
c sin  4   3   a2 sin  2  a2 2 cos  2  b32 cos3  c4 2 cos 4 

 
c cos 4   4   a2 cos 2  a2 2 sin  2  b32 sin 3  c4 2 sin  4 
Solving for  3  3 ,  4  4
A  c sin  4
--&gt;
CD  AF
3 
AE  BD where
CE  BF
4 
AE  BD
B  b sin 3
C  a 2 sin  2  a2 2 cos 2  b32 cos3  c4 2 cos 4
D  c cos  4
E  b cos3
F  a 2 cos  2  a2 2 sin  2  b32 sin 3  c4 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 j2  be j3  ce j3  de j1  0


Differentiate with time twice (d and 3 are variables)
 aj  a  e
2
2

2
j 2


 bj3  b32 e j3  d  0
Inserting Euler form
aj2 (cos  2  j sin  2 )  a2 2 (cos 2  j sin  2 )  bj3 (cos 3  j sin 3 )  b32 (cos 3  j sin 3 )  d  0

Separate real and imaginary parts
 a2 sin  2  a2 2 cos 2  b3 sin 3  b32 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
--&gt;
a 2 cos  2  a2 2 sin  2  b32 sin 3
3 
b cos3
d   a 2 sin  2  a2 2 cos  2  b 3 sin 3  b32 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 j2

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
 j2  p2 je j2  V p _ tran  V p _ slip
VP  pe
AP  
pe j2  p2 2e j2  2 p2 je j2  p2 je j2 Ap
 Ap _ slip  Ap _ normal  Ap _ Coriolis  Ap _ tan gential

Physically interpret each terms in vector diagram
&lt;--
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 j4  0
aj2 e j2  bj  e j   be
4
Acceleration vector loop equation
 a
2




je j2  a2 2 e j2    b  b 2 e j  b  2b je j    c 4 je j4  c4 2 e j4   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
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 j2  pe j (3 3 )
Position U: RU  ue j (4  4 )
Example: Acceleration of point P
RP  ae j2  pe j (3 3 )
AP  a 2 je j2  a2 2 e j2  p 3 je j (3 3 )  p32 e j (3 3 )
 a 2 j (cos  2  j sin  2 )  a2 2 (cos  2  j sin  2 )  p 3 j (cos(3   3 )  j sin(3   3 ))  p32 (cos(3   3 )  j sin(3   3 ))
  a 2 sin  2  a2 2 cos  2  p 3 sin(3   3 )  p32 cos(3   3 ) 
 j  a 2 cos  2  a2 2 sin  2  p 3 cos(3   3 )  p32 sin(3   3 ) 
APX  a 2 sin  2  a2 2 cos  2  p 3 sin(3   3 )  p32 cos(3   3 )
APY  a 2 cos  2  a2 2 sin  2  p 3 cos(3   3 )  p32 sin(3   3 )
```