-1-
Content
-2-
Orthonormal basis of 3D space: Definition
A well known Orthonormal basis of a
3D space consists of three unit vectors
along x, y, and z axes, denoted as i, j,
k, respectively, or alternatively as e1, e2,
and e3, or ex, ey, and ez.
Kinematics of planar mechanisms
using Orthonormal basis
 Orthonormal basis of 3D space
• Definition
• Scalar product
• Vector product
z

k
O
The basis are orthogonal:
 

 

i ^ j , j ^ k ,k ^ i
 Revision: Motion of points on the body
 Example
y

i
and normal:



i = j = k =1
• Crank-slider mechanism
• Planar quadrilateral linkage

j
x
For any 3D vector, the following
expression can be uniquely written:




v = vx i + vy j + vz k
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-3-
Orthonormal basis of 3D space: Scalar product
Definition:
 
a ⋅ b = ab cos q
Some properties:
where
 
q = (a , b )
Definition:
q
q
 
 
cos q > 0 cos q < 0
a ⋅b = b ⋅a
  
   
a ⋅ (b + c ) = a ⋅ b + a ⋅ c
For arbitrary vectors:



 
  



a ⋅ (rb + sc ) = ra ⋅ b + sa ⋅ c v = x i + y j + z k
1
1
1
1





(ra ) ⋅ (sb ) = rsa ⋅ b

v 2 = x 2 i + y 2 j + z 2k
For unit vectors:
 
 
 
ß
i ⋅ i = 1; j ⋅ j = 1; k ⋅ k = 1  
v1v 2 = x 1x 2 + y1y 2 + z 1z 2
 
 
 
i ⋅ j = 0; j ⋅ k = 0; k ⋅ i = 0
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-4-
Orthonormal basis of 3D space: Vector product

 
c = a ´b ,
c = ab sin q
 
q = (a , b )
Some properties:
 
 
a ´ b = -b ´ a
 

   
a ´ (b + c ) = a ´ b + a ´ c




 
a ´ (rb ) = (ra ) ´ b = r (a ´ b )
For unit vectors:
 
  
  

i ´ i = 0; j ´ j = 0; k ´ k = 0
  

 
  
i ´ j = k ; j ´k = i ;i ´ j = k
  
 
  

j ´ i = -k ; k ´ j = -i ; i ´ k = - j
i, j, k must follow the
right-hand rule!!!
Positive sense:

k
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME

i

j
-5-
Orthonormal basis of 3D space: Vector product (cont.)
For arbitrary vectors:
Or, calculate a determinant:




v1 = x 1i + y 1 j + z 1k




v 2 = x 2 i + y 2 j + z 2k

i
 
v1 ´ v 2 = x 1
x2
ß

j
y1
y2
Velocity of a point on the body

k
z1
z2


 
vP = vA + w ´ u


= vA + vPA,

 
vPA = w ´ u
P

 
vPA = w ´ u

^u
A
PA

vPA
lies on the plane of
the plate
w
y

rP

u
A

rA
O
x


  
 
a P = a A + a ´ u + w ´ (w ´ u )

a
w
= a A + a PA
+ a PA
a
 
aPA
= a ´u
w

 
aPA
= w ´ (w ´ u )
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-7-

vP
P
Acceleration of a point on the body






 
v1 ´ v 2 = (x 1i + y 1 j + z 1k ) ´ (x 2i + y 2 j + z 2k )
 
 
 
= x 1y 2 (i ´ j ) + x 1z 2 (i ´ k ) + y 1x 2 ( j ´ i ) +
 
 
 
+y 1z 2 ( j ´ k ) + z 1x 2 (k ´ i ) + z 1y 2 (k ´ j )



= (y 1z 2 - z 1y 2 )i + (z 1x 2 - x 1z 2 ) j + (x 1y 2 - y 1x 2 )k
3. Motion of points on the body: Relative motion analysis





Note: for planar motion w ^ u
v = v +v
-6-
3. Motion of points on the body: Relative motion analysis

vA
 P 
u
v
A
Note: for planar motion

 
w, a ^ u



a P = a A + a PA
t
 
a PA
= a ´ u,

^ u;

t
n
a PA = aPA
+ a PA
n


 
a PA
= w ´ (w ´ u ) = -w 2u


 

aP = a A + a ´ u - w 2u
A
angular acceleration is
counter-clockwise
t
aPA
A
angular acceleration is
clockwise
P
P

u
a
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-8-
3. Motion of points on the body: Relative motion analysis
n
aPA

aA
a

u
A
n
a PA
t
aPA

aA
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-9-
Example 3. Crank-slider mechanism
Crank-slider mechanism OAB at the instant considered is at a position such
that OA is perpendicular to AB, OA forms an angle of 60 with the horizontal
plane. Given that crank OA rotates at a constant angular velocity .
a) Determine and draw the velocities of points A A
and B, and angular velocity of AB.
r
b) Determine the acceleration of point B.
0
Solution
Determination of velocities (using vector-equation)
l
  
{i , j , k }
y




wOA = w0k ,
aOA = 0k ,



OA = r cos 60i + r sin 60 j


= 21 ri + 23 rj
 


l = AB = l cos 30i - l sin 30 j


= 23 li - 21 lj
B
Preliminary motion analysis
How the bodies move? Which motions are
known?
l = 3r
vA
A

j
l
r
0
60o

i
O
vB


wAB = wAB k ,


Need to determine:
vB = vB i ,
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME


v B = vB i ,




vB /A = wAB ´ AB = wAB k ´ (l


= 23 l wAB j + 21 l wAB i
= ( 21 l wAB -
3
2
3
2
3
2

j

rj )
(
A
l
r
0
Calculation
60o

i
O


3
i - l 21 j )
2
vB
vB = ( 21 l wAB -



r w0i + 23 l wAB j + 21 l wAB i


r w0 )i + ( 21 r w0 + 23 l wAB )j
0 = ( 21 r w0 +
x
B
3
2
3
2
r w0 )
l wAB )
wAB = -w0 / 3
vB = -2r w0 / 3
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME


aAB = aAB k


aB = aB i ,
-12-
Example 3. Crank-slider mechanism
Applying formula
 





a B = a A + aAB ´ l + wAB ´ wAB ´ l




2
= a A + aAB ´ l - wAB
l
l = 3r
vA





vA = wOA ´OA = w0k ´ ( 21 ri +


= 12 r w0 j - 23 r w0i


vB i = 21 r w0 j -
y
x
B
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-11-
Example 3. Crank-slider mechanism



Applying formula 
vB = vA + wAB ´ AB
  
{i , j , k }
Setting a Cartesian coordinate system Oxyz with unit vectors
Presenting all vectors with
60
O
-10-
Example 3. Crank-slider mechanism





a A = wOA ´ (wOA ´OA) = -w02OA


= - 21 r w02i - 23 r w02 j
y
)

j
0
O
l = 3r
A
aA
l
60o

i


aB = aB i ,
aB
x
B


aAB = aAB k






a Bt /A = aAB ´ AB = aAB k ´ (l 23 i - l 21 j )


= 23 l aAB j + 21 l aAB i




2
2
a Bn /A = -wAB
AB = -wAB
(l 23 i - l 21 j )


2
2
3
1
= -l wAB
i + l wAB
j
2
2
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-13-
Example 3. Crank-slider mechanism
y
Putting all terms into this eq.





2
a B = aA + aAB ´ l - wAB
l

j
yields
3
2

r w02 j +
= (- 21 r w02 + 21 l aAB

i
3
2
rw +
3
2
x
aB
B




2
2
3
1
l aAB j + 21 l aAB i - l wAB
i + l wAB
j
2
2


2
1
)j
- l w 23 )i + (- 23 r w02 + 23 l aAB + l wAB
2
3
2
2
AB
2
a B = - 21 r w02 + 21 l aAB - l wAB
2
0
l
60o
O


aB i = - 21 r w02i -
0=-
aA
0
Planar quadrilateral linkage OABO1 at
instant considered has the configuration as
shown in the figure (OA is vertical, AB
horizontal). Given OA = r, AB = l, and
crank O1B is having an angular velocity 0
and zero angular acceleration.
Determine the angular velocity and angular
acceleration of crank OA.
l = 3r
A
l aAB + l w
aAB =
2
3r w02 - l wAB
3
3l
2
a B = - 21 r w + 21 l aAB - l wAB
OA = r ,
= ..
-15-
B
O



rOA = OA = rj ,




rO B = O1B = -r .i + r .j ,
1
 


rAB = l = AB = l .i
Calculation


vB

i
45
O1
Need to determine:
wAB , aAB , wOA , aOA
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
AB = l ,
O1B = r 2
x
-16-
Example 4. Planar quadrilateral linkage



Applying formula 
vB = vA + wAB ´ AB
  
k = i ´j
y
j
O1
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
Example 4. Planar quadrilateral linkage
 A
vA 
45
O
Determination of lengths
3
2
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
  
Presenting all vectors with {i , j , k }




wO B = w0k ,
aO B = 0k ,
1
1




wAB = wAB k , aAB = aAB k ,




wOA = wOAk , aOA = aOAk ,

Preliminary motion analysis
How the bodies move? Which motions are
known?
2
0
Setting a Cartesian coordinate   
system Oxyz with unit vectors {i , j , k }
B
A
Solution
3
2
2
1
AB 2
-14-
Example 4. Planar quadrilateral linkage
y
 A
vA 
B


j


vB
vB = wO B ´O1B
1


O
= w0k ´ (-r .i + r .j )

  
i
i
j k


= 0 0 w0 = -r w0 j - r w0i
-r r 0






 


vB /A = wAB ´ AB = wAB k ´ l .i
vA = wOA ´OA = wOAk ´ rj

 

 
i j
k
i j
k


= 0 0 wAB = l wAB j
= 0 0 wOA = -r wOAi
l 0
0
0 r
0

45
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
O1
x
-17-
Example 4. Planar quadrilateral linkage
  
Substituting vB , vA , vAB
into
B
 A
vA 




vB = vA + wAB ´ AB
j
O




-r w0 j - r w0i = -r wOAi + l wAB j
-r w0 = -r wOA ,
Applying formula





2
a B = a A + aAB ´ l - wAB
l
y


vB
45

i





a B = a Bn = -w02O1B = r w02i - r w02 j
 


a A = -w02OA + aOA ´OA


2
= -r wOA
j - r aOAi
x
wOA = w0
wAB = -r w0 / l

j
O



vA = -r wOAi = -r w0i



vB /A = l wAB j = -r w0 j
y



2
r w02i - r w02 j = -(l wAB
+ r aOA )i

2
+(l aAB - r wOA
)j
A

j
O
t
aBA
n
aBA

aAn
B

i

i
B


aB  aBn

45


aOA = aOAk ,


aAB = aAB k
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
-19-

aAt

aAn






2
2
r w02i - r w02 j = -r wOA
j - r aOAi + l aAB j - l wAB
i
Example 4. Planar quadrilateral linkage




2
r w02i - r w02 j = -r wOA
j - r aOAi


2
+l aAB j - l wAB
i
t
aBA
n
aBA
A



2
2
a Bn /A = -wAB
l = -l wAB
i
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
2
-r w02 = (l aAB - r wOA
)

aAt






aBt /A = aAB ´ AB = aAB k ´ (l .i ) = l aAB j
-r w0 = l wAB
2
r w02 = -(l wAB
+ r aOA )
y
Calculation
O1
-18-
Example 4. Planar quadrilateral linkage


aB  aBn

45


aOA = aOAk ,


aAB = aAB k
2
aOA = -l wAB
/ r - w02
2
aAB = (r wOA
- r w02 ) / l
Nguyễn Quang Hoàng-Department of Applied Mechanics-SME
O1
x
O1
x