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Presented by-
Sameer Vishwakarma
M.Tech. (Production)
III SEM.
Submitted to-
Dr. Manoj Kumar Agrawal
Associate Professor
Institute of Engineering and Technology
Department of Mechanical Engineering
Basic Joints
Spherical Joint
3 DOF ( Variables - 1, 2, 3)
Revolute Joint
1 DOF ( Variable - )
Prismatic Joint
1 DOF (linear) (Variables - d)
We are interested in two kinematics topics
Forward Kinematics (angles to position)
What you are given: The length of each link
The angle of each joint
What you can find: The position of any point
(i.e. it’s (x, y, z) coordinates
Inverse Kinematics (position to angles)
What you are given: The length of each link
The position of some point on the robot
What you can find: The angles of each joint needed to obtain
that position
Quick Math Review
Dot Product:
Geometric Representation:
A
B
θ
cosθBABA 
Unit Vector
Vector in the direction of a chosen vector but whose magnitude is 1.
B
B
uB 






y
x
a
a






y
x
b
b
Matrix Representation:
yyxx
y
x
y
x
baba
b
b
a
a
BA 












B
Bu
Quick Matrix Review
Matrix Multiplication:
An (m x n) matrix A and an (n x p) matrix B, can be multiplied since the number of columns of A
is equal to the number of rows of B.
Non-Commutative Multiplication
AB is NOT equal to BA
   
   



















dhcfdgce
bhafbgae
hg
fe
dc
ba
Matrix Addition:
   
   



















hdgc
fbea
hg
fe
dc
ba
Basic Transformations
Moving Between Coordinate Frames
Translation Along the X-Axis
N
O
X
Y
Px
VN
VO
Px = distance between the XY and NO coordinate planes






 Y
X
XY
V
V
V 





 O
N
NO
V
V
V 






0
P
P x
P
(VN,VO)
Notation:
N
X
P
VN
VO
Y O

NO
O
N
XXY
VP
V
VP
V 




 

Writing in terms ofXY
V NO
V
X
N
VN
VO
O
Y
Translation along the X-Axis and Y-Axis








 O
Y
N
XNOXY
VP
VP
VPV







Y
xXY
P
P
P


































oV
nV
θ)cos(90V
cosθV
sinθV
cosθV
V
V
V NO
NO
NO
NO
NO
NO
O
N
NO
NO
V
o
n Unit vector along the N-Axis
Unit vector along the N-Axis
Magnitude of the VNO vector
Using Basis Vectors
Basis vectors are unit vectors that point along a coordinate axis
NVN
VO
O
n
o
Rotation (around the Z-Axis)
X
Y
Z
X
Y

V
VX
VY






 Y
X
XY
V
V
V 





 O
N
NO
V
V
V
 = Angle of rotation between the XY and NO coordinate axis
X
Y

V
VX
VY

Unit vector along X-Axis
x
xVcosαVcosαVV NONOXYX

NOXY
VV 
Can be considered with respect to
the XY coordinates or NO coordinates
V
x)oVn(VV ONX

(Substituting for VNO using the N and O components of the
vector)
)oxVnxVV ONX
 ()(
))
)
(sinθV(cosθV
90))(cos(θV(cosθV
ON
ON


Similarly….
yVα)cos(90VsinαVV NONONOY

y)oVn(VV ONY

)oy(V)ny(VV ONY

))
)
(cosθV(sinθV
(cosθVθ))(cos(90V
ON
ON


So….
)) (cosθV(sinθVV ONY

)) (sinθV(cosθVV ONX







 Y
X
XY
V
V
V
Written in Matrix Form











 






 O
N
Y
X
XY
V
V
cosθsinθ
sinθcosθ
V
V
V
Rotation Matrix about the z-axis
X1
Y1

VXY
X0
Y0
VNO
P











 












 O
N
y
x
Y
X
XY
V
V
cosθsinθ
sinθcosθ
P
P
V
V
V
(VN,VO)
In other words, knowing the coordinates of a point (VN,VO) in some coordinate
frame (NO) you can find the position of that point relative to your original
coordinate frame (X0Y0).
(Note : Px, Py are relative to the original coordinate frame. Translation followed by rotation is different than rotation
followed by translation.)
Translation along P followed by rotation by 











 












 O
N
y
x
Y
X
XY
V
V
cosθsinθ
sinθcosθ
P
P
V
V
V
HOMOGENEOUS REPRESENTATION
Putting it all into a Matrix



















 























1
V
V
100
0cosθsinθ
0sinθcosθ
1
P
P
1
V
V
O
N
y
x
Y
X



















 












1
V
V
100
Pcosθsinθ
Psinθcosθ
1
V
V
O
N
y
x
Y
X
What we found by doing a translation
and a rotation
Padding with 0’s and 1’s
Simplifying into a matrix form









 

100
Pcosθsinθ
Psinθcosθ
H y
x
Homogenous Matrix for a Translation in XY plane,
followed by a Rotation around the z-axis

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Robot kinematics

  • 1. Presented by- Sameer Vishwakarma M.Tech. (Production) III SEM. Submitted to- Dr. Manoj Kumar Agrawal Associate Professor Institute of Engineering and Technology Department of Mechanical Engineering
  • 2. Basic Joints Spherical Joint 3 DOF ( Variables - 1, 2, 3) Revolute Joint 1 DOF ( Variable - ) Prismatic Joint 1 DOF (linear) (Variables - d)
  • 3. We are interested in two kinematics topics Forward Kinematics (angles to position) What you are given: The length of each link The angle of each joint What you can find: The position of any point (i.e. it’s (x, y, z) coordinates Inverse Kinematics (position to angles) What you are given: The length of each link The position of some point on the robot What you can find: The angles of each joint needed to obtain that position
  • 4. Quick Math Review Dot Product: Geometric Representation: A B θ cosθBABA  Unit Vector Vector in the direction of a chosen vector but whose magnitude is 1. B B uB        y x a a       y x b b Matrix Representation: yyxx y x y x baba b b a a BA              B Bu
  • 5. Quick Matrix Review Matrix Multiplication: An (m x n) matrix A and an (n x p) matrix B, can be multiplied since the number of columns of A is equal to the number of rows of B. Non-Commutative Multiplication AB is NOT equal to BA                            dhcfdgce bhafbgae hg fe dc ba Matrix Addition:                            hdgc fbea hg fe dc ba
  • 6. Basic Transformations Moving Between Coordinate Frames Translation Along the X-Axis N O X Y Px VN VO Px = distance between the XY and NO coordinate planes        Y X XY V V V        O N NO V V V        0 P P x P (VN,VO) Notation:
  • 8. X N VN VO O Y Translation along the X-Axis and Y-Axis          O Y N XNOXY VP VP VPV        Y xXY P P P
  • 9.                                   oV nV θ)cos(90V cosθV sinθV cosθV V V V NO NO NO NO NO NO O N NO NO V o n Unit vector along the N-Axis Unit vector along the N-Axis Magnitude of the VNO vector Using Basis Vectors Basis vectors are unit vectors that point along a coordinate axis NVN VO O n o
  • 10. Rotation (around the Z-Axis) X Y Z X Y  V VX VY        Y X XY V V V        O N NO V V V  = Angle of rotation between the XY and NO coordinate axis
  • 11. X Y  V VX VY  Unit vector along X-Axis x xVcosαVcosαVV NONOXYX  NOXY VV  Can be considered with respect to the XY coordinates or NO coordinates V x)oVn(VV ONX  (Substituting for VNO using the N and O components of the vector) )oxVnxVV ONX  ()( )) ) (sinθV(cosθV 90))(cos(θV(cosθV ON ON  
  • 12. Similarly…. yVα)cos(90VsinαVV NONONOY  y)oVn(VV ONY  )oy(V)ny(VV ONY  )) ) (cosθV(sinθV (cosθVθ))(cos(90V ON ON   So…. )) (cosθV(sinθVV ONY  )) (sinθV(cosθVV ONX         Y X XY V V V Written in Matrix Form                     O N Y X XY V V cosθsinθ sinθcosθ V V V Rotation Matrix about the z-axis
  • 13. X1 Y1  VXY X0 Y0 VNO P                           O N y x Y X XY V V cosθsinθ sinθcosθ P P V V V (VN,VO) In other words, knowing the coordinates of a point (VN,VO) in some coordinate frame (NO) you can find the position of that point relative to your original coordinate frame (X0Y0). (Note : Px, Py are relative to the original coordinate frame. Translation followed by rotation is different than rotation followed by translation.) Translation along P followed by rotation by 
  • 14.                           O N y x Y X XY V V cosθsinθ sinθcosθ P P V V V HOMOGENEOUS REPRESENTATION Putting it all into a Matrix                                             1 V V 100 0cosθsinθ 0sinθcosθ 1 P P 1 V V O N y x Y X                                  1 V V 100 Pcosθsinθ Psinθcosθ 1 V V O N y x Y X What we found by doing a translation and a rotation Padding with 0’s and 1’s Simplifying into a matrix form             100 Pcosθsinθ Psinθcosθ H y x Homogenous Matrix for a Translation in XY plane, followed by a Rotation around the z-axis