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Intuitive Kinematics – Converting Between Forward and Reverse Definitions of Space. Lecture Series 2 ME 4135 R. R. Lindeke. Intuitive Kinematics for Robot Manipulators. Defining the concept of the Kinematic Solution Finding Kinematic Solutions for POSITIONAL issues (only!) - PowerPoint PPT Presentation
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Intuitive Kinematics – Converting Between Forward and Reverse Definitions of Space
Lecture Series 2ME 4135 R. R. Lindeke
Intuitive Kinematics for Robot Manipulators Defining the concept of the Kinematic
Solution Finding Kinematic Solutions for
POSITIONAL issues (only!) Cartesian Manipulators Cylindrical Manipulators Spherical Manipulators Articulating Manipulators SCARA and Other Redundant
Manipulators
Intuitive Kinematics for Robot Manipulators Forward Kinematic Solutions:
Given the settings on each Joint (q1, q2, , qi) of the manipulator …
Determine the End Position (Xe,Ye, Ze)base achieved for the given structural size
Inverse Kinematic Solutions: Given Structural Size and an End (or Target)
Position (Xe,Ye, Ze)base … Determine the values for each Joint (q1, q2, , qi)
that can place the Manipulator there But, Note, this solution MAY NOT BE UNIQUE!
Cantilevered Cartesian Robot
P-P-P Configuration
J1
J3
J2
X0
Y0
Z0
Gantry Cartesian RobotP-P-P Configuration
X0Y0
Z0
J1
J2J3
FKS & IKS for a Cartesian Device In the Forward Sense: On the Gantry
J1 was at 435 cm J2 was at 283 cm J3 was at 199 cm and there is a COLLAPSE LENGTH of 75,
50, & 50 cm respectively Where is the End in the ‘Null (base)
Space’?
FKS -- Cartesian
Xe is J1 + Cl1 = 435 + 75 = 510 Ye is 283 + 50 = 333 Ze is 199 + 50 = 249 Exercise Care in the ordering of the
Joints as they relate to the Base Frame definition of Space!!!
Doing the IKS Given you want the End Position of:
(510, 333, 249) and the Collapse Lengths are as seen above (75, 50, & 50 respectively)
J1 is in the X0 direction therefore, J1 = 510 – 75 = 435 (as expected!) Similarly for J2 & J3 (which are 283 and 199
respectively) – again as expected Here because of the directions the Joint motions
follow and how the Base Axes are defined: J1 in the direction of X0; J2 in Y0; J3 in Z0 – but
be careful as we move deeper into robotics!
Cylindrical Robot Work EnvelopeP-R-P//R-P-P Configuration
Cylindrical Robot
Developing a FKS (model):
Given , Z & R Compute End
Position in terms of X0, Y0 and Z0 (Xe,Ye, Ze)
R
X0
Y0
Z0
FKS for Cylindrical Manipulator
( )*
( )*
( )
mach
mach
mach
mach R
mach R
mach Z
Xe R Cl Cos
Ye R Cl Sin
Ze Z Cl
IKS for Cylindrical Manipulator Here, before we go on let me state this:
Sine and Cosine inverses lead to ambiguous angles (they repeat each semi-circle) since they are built from a ratio of a signed over an unsigned vector (Y/R or X/R )
We MUST use inverse Tangent solutions to remove the ambiguity since tangent is built from a ratio of signed vectors (Y/X)
Cylindrical Robot
Doing IKS – given a value for:
X, Y and Z of End Compute , Z and
R
R
X0
Y0
Z0
IKS Cylindrical Manipulator2 2
3
*( / )*
2( , ):
ATan2 is a special form of InverseTangent that places the angle in the proper quadrant (I to IV) depending on the signs of Xe and Ye (the
R Xe Ye ClR SinYe Xe TanR Cos
ATan Xe Yewhere
2
denominatorand numerator of the ratio respectively)
machZ Ze Cl
Computing ATan2 Angles Atan2(Xvalue,Yvalue) is a special form of Tan-1 but computed to
retain quadrant identity Consider the 4-cases of: X = 8 and Y = 12
Atan2(8,12) = Tan-1(12/8) = 56.31 This is a 1st Quadrant angle!
Atan2(-8,12) = 90 + Tan-1(8/12) = 90 + 33.69 = 123.69 (alternatively it is 180 - Tan-1(12/8) = 180 – 56.31 = 123.69) This is a 2nd quadrant angle and not the -56.31 value that you
find when computing Atan with your calculator! Atan2(-8,-12) =180 + Tan-1(12/8) =180 + 56.31 = 236.31
This is a 3rd quadrant angle! Atan2(8,-12) = 270 + Tan-1(8/12) = 270 + 33.69 = 303.69
(alternatively it is 360 - Tan-1(12/8) = 360 – 56.31 = 303.69) This is a 4th quadrant angle which is the same as your
calculator gives you (-56.31)!
Spherical Robot WorkspaceR-R-P Configuration
Spherical Robot
Developing a FKS (model):
Given , & R Compute Xe, Ye, &
Ze
R
X0
Y0
Z0
FKS Spherical Manipulator
0 0R' = projection of Radial Arm to X --Y plane' *
'*'*
*
given
given
R R Cos
Xe R CosYe R SinZe R Sin
Spherical Robot
Developing a IKS (model):
Given Xe, Ye, & Ze Compute , & R
R
X0
Y0
Z0
IKS, Spherical Manipulator
2 2 23
2 2
2 2
R = XeAgain, for Angles we desire :
''
tan 2( , )
'
tan 2( ', ):
':
tan 2 ,
: can be thought of
Ye Ze Cl
Ye R Sin TanXe R Cos
A Xe YeZe RSin TanR RCos
A R Zewhere
R Xe Yeso
A Xe Ye Ze
note
Tangent Inverses
2 as
2-Link Articulating Arm Manipulator
3
1
2
L2
L1
• This Machine Rotates about Z0 Axis (1)
• 2 is measured RELATVE to the base plane
• 3 is measured RELATVE to L1 not the base plane
• All Joint Angles are Right Hand Rule Based
Z0
X0Y0
FKS 2-Link A. ARM Manipulator
1 1 2 2 2 3
1 1 2 2 2 3
1 2 2 2 3
* * *
* * *
* *
Xe Cos L Cos L Cos
Ye Sin L Cos L Cos
Ze L Sin L Sin
2-Link Articulating Arm Manipulator
3
12
L2
L1
Z0
X0Y0
IKS 2-Link A. Arm All angles defined as ATan2 (using
target End Position & Link Lengths) Requires Construction Lines!! First Solve for 1 = ATan2(Xe, Ye) Then the Tilt Angles (3 & 2) in
reversed Order (as stated)! Solution Indicates 2 acceptable
configurations for the Arm: ‘Elbow UP’ and ‘Elbow DOWN’
IKS 2-Link A. Arm1
1
0 0
3 2
tan 2( , )Rotate Arm to Position andDefine a Plane normal to X -Y andContaining the Arm LinksDefine a Vector R from origin of newPlane to End of ArmSolve for and
A Xe Ye
IKS Q2 & Q3
2
3
a
IKS 2-Link Art. Arm (3)
2 2 21 2 1 2 3
2 2 2 2
3 3
3
2 2 21 2
321
33
3
2 23 3
2
3
23
Using Law of Cosines:
2 (180 ):
:(180 )
solving for COS :
2
:
1 1
1
2 , 1
R L L L L Coswhere
R Xe Ye Ze
andCos Cos
R L LCos DL L
SinTan Cosand
Sin Cos D
DTan D
ATan D D
IKS 2-Link Art. Arm (2)
00
1
2
2 2
2 2 22 1 1
2 2 21 2
1
Lets define the angle from the X -Y planeto Vector R and as the angle between L and R
:
tan 2 ,
Solving for :Law of Cosines:
2 ( )
( ) 2Law o
Therefore
A Xe Ye Ze
L R L RLCos
R L LCos RL
a
a
a
a
a
3
2
3 3
2 3
f Sines:( ) (180 )
:(180 )
( )
Sin SinL R
hereSin Sinso
L SinSin R
a
a
IKS 2-Link A. Arm (2) cont.
2 3
2 2 21 2
1
1 2 32 2 2
1 2
2 2 21 2 1 2 3
( )( ) ( )
( )
2:2( )
2 , 2
SinTan CosL Sin
RTanR L L
RLsimplifying
L L SinTanR L L
ATan R L L L L Sin
aa a
a
a
a
IKS 2-Link A. Arm (2) cont.
2 2 2 2 22 1 2 1 2 3
23
2 2 2 2 2 22 1 2 1 2
2 2 21 2
21
:
tan 2 , 2 , 2
Re
1
tan 2 , 2 , 2 * 1
:
2
Finally
A Xe Ye Ze ATan R L L L L Sin
membering
Sin D
A Xe Ye Ze ATan R L L L L D
where
R L LDL L
a
SCARA Manipulator – an over specified Planer Articulating Arm
FKS & IKS for this over specified Arm – Only 2 “State Equations” Exist for 3 Variables
In a Forward Sense solution is reasonable
In an Inverse sense the above statement indicates the existence of an infinite number of possible solutions
FKS: Given 1,2,3 find Xe and Ye
IKS: Given Xe and Ye (& L1, L2 and L3) find: 1,2,3
FKS 3-Link Planer A. Arm
y1
y2
y3
x1 x2 x3
1
2
3
Focusing on the FKS Project each link to the two axes (X & Y) Xe is found by summing X-projected
lengths of each link Ye found by summing Y-projected lengths
of the links Example:
Link 1 to X: L1*Cos(1) Link 1 to Y: L1*Sin(1)
FKS Continuing X2 is projection of L2 – to the X axis therefore it
requires what Projection Factor? Sure, Cos(1+ 2) Y2 is projection of L2 to the Y axis so it is equal
to: L2* Sin(1+ 2) Finally:
Xe = L1*Cos(1) + L2*Cos(1+ 2) + L3*Cos(1+ 2+ 3)
Ye = L1*Sin(1) + L2*Sin(1+ 2) + L3*Sin(1+ 2+ 3)
What about the IKS It’s a 2 Step Process Requires a Parameterization of one of
the Joint Angles – This step will establish the acceptable limits of the solution space
The Parameterized Joint is said to set the BOUNDS for the solution space
We select Joint 1 as the one to be parameterized
3-Link Planer Arm IKS – Step 1
1
2’
Note: L2’ is the sum of L2 + L3 formed by freezing 3 at 0˚
1 Is Solved as Above It is the lower angle of a 2-link
Articulating Arm The 2 solutions found for 1 thus form
the Upper and Lower Bounds for the solution space
Now pick one within the range Using this angular value, “Transform”
the solution space up Link L1 and ‘unfreeze’ 3
IKS Step 2:’Redefine’ Space at end of L1 and Re-apply 2-link method in the Transformed space
1,picked
Homework Assignment: Complete the IKS solution for a 3-link
planar articulating arm Consider and complete the FKS and
IKS solution for a “Planer” P-R-P device