1. Robotic Kinematics – the Inverse Kinematic Solution
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Kinematics and Mechatronics
Dr. R. Lindeke
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2. FKS vs. IKS
In FKS we built a tool for finding end frame geometry from Given Joint data:
In IKS we need Joint models from given End Point Geometry:
Joint Space
Cartesian Space
Joint Space
Cartesian Space
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3. It a more difficult problem because:
So this IKS Problem is Nasty (in Mathematics this is considered a Hard Modeling Issue)
It a more difficult problem because:
The Equation set is “Over-Specified”:
12 equations in 6 unknowns
Space can be “Under-Specified”:
Planer devices with more joints than 2
The Solution set can contain Redundancies:
Multiple solutions
The Solution Sets may be un-defined:
Unreachable in 1 or many joints
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4. But the IKS is VERY Useful – some Uses of IKS include:
Building Workspace Maps
Allow “Off-Line Programming” solutions
The IKS allows the engineer to equate Workspace capabilities with Programming realities to assure that execution is feasible (as done in ROBCAD & IGRIP)
The IKS Aids in Workplace Design and Operational Simulations
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5. Doing a Pure IKS solution: the R Manipulator
R Frame Skeleton (as DH Suggest!)
Same Origin Point!
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6. LP Table and Ai’s Frames Link Var  d a  S C S  C  0  1 1 R
 + 90 90 C1
-S1
1  2 2 P
d2 + cl2
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7. FKS is A1*A2:
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8. Forming The IKS:
In the Inverse Problem, The RHS MATRIX is completely known (perhaps from a robot mapped solution)! And we use these values to find a solution to the joint equations that populate the LHS MATRIX
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9. Forming The IKS: Form a ratio to build Tan():
Examining these two matrices
n, o, a and d are “givens” in the inverse sense!!!
(But typically we want to build general models leaving these terms unspecified)
Term (1, 4) & (2,4) on both sides allow us to find an equation for :
(1,4): C1*(d2+cl2) = dx
(2,4): S1*(d2+cl2) = dy
Form a ratio to build Tan():
S1/C1 = dy/ dx
Tan  = dy/dx
 = Atan2(dx, dy)
If 2 Matrices are Equal then EACH and EVERY term is also Uniquely equal as well!
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10. After  is found, back substitute and solve for d2:
Forming The IKS:
After  is found, back substitute and solve for d2:
(1,4): C1*(d2+cl2) = dx
Isolating d2: d2 = [dx/Cos1] - cl2
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11. Alternative Method – “doing a pure inverse approach”
Form A1-1 then pre-multiply both side by this ‘inverse’
Leads to: A2 = A1-1*T0ngiven
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12. Simplifying & Solving:
Selecting and Equating (1,4)
0 = -S1*dx + C1*dy
Solving: S1*dx = C1*dy
Tan() = (S1/C1) = (dy/dx)
 = Atan2(dx, dy) – the same as before
Selecting and Equating (3,4) -- after back substituting  solution
d2 + cl2 = C1*dx + S1*dy
d2 = C1*dx + S1*dy - cl2
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13. Performing IKS For Industrial Robots: (a more involved problem)
First let’s consider the concept of the Spherical Wrist Simplification:
All Wrist joint Z’s intersect at a point
The ‘n Frame’ is offset from this Z’s intersection point at a distance dn (the hand span) along the a vector of the desired solution (3rd column of desired orientation sub-matrix!) which is the z direction of the 3rd wrist joint
This follows the DH Algorithm design tools as we have learned them!
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14. Performing IKS – with a Spherical Wrist
We can now separate the POSE effects:
ARM joints
Joints 1 to 3 in a full function manipulator (without redundant joints)
They function to maneuver the spherical wrist to a target POSITION related to the desired target POSE
WRIST Joints
Joints 4 to 6 in a full functioning spherical wrist
Wrist Joints function as a primary tool to ORIENT the end frame as required by the desired target POSE
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15. Performing IKS: Focus on Positioning
We will define a point (called the WRIST CENTER) where all 3 z’s intersect as:
Pc = [Px, Py, Pz]
We find that this position is exactly: Pc = dtarget - dn*a
Px = dtarget,x - dn*ax
Py = dtarget,y - dn*ay
Pz = dtarget,z - dn*az
Note: dn is the so called ‘Handspan’ (a CONSTANT)
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16. Treating the Arm Types (as separate robot entities):
Cartesian (2 types)
Cantilevered
Gantry
Cylindrical
Spherical (w/o d2 offset)
2 Link Articulating (w/o d2 offset)
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17. Cantilevered Cartesian RobotJ2 J3 X0
P-P-P Configuration Y0 Z0 J1
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18. Gantry Cartesian Robot
P-P-P Configuration X0 Z0 Y0 J3 J2 J1
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19. Cylindrical Robot Doing IKS – given a value for: X, Y and Z of End
Compute , Z and R Z0 R Y0 X0
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20. Spherical Robot  Developing a IKS (model): Given Xe, Ye, & Ze
Compute ,  & R R Z0 Y0 X0
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21. 2-Link Articulating Arm ManipulatorZ0 3 L2 L1 X0 Y0 2 1
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22. Focusing on the ARM Manipulators in terms of Pc:
Prismatic:
q1 = d1= Pz (its along Z0!) – cl1
q2 = d2 = Px or Py - cl2
q3 = d3= Py or Px - cl3
Cylindrical:
1 = Atan2(Px, Py)
d2 = Pz – cl2
d3 = Px/C1 – cl3 {or +(Px2 + Py2).5 – cl3}
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23. Focusing on the ARM Manipulators in terms of Pc:
Spherical:
1 = Atan2(Px, Py)
2 = Atan2( (Px2 + Py2).5 , Pz)
D3 = (Px2 + Py2 + Pz2).5 – cl3
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24. Focusing on the ARM Manipulators in terms of Pc:
Articulating:
1 = Atan2(Px, Py)
3 = Atan2(D, (1 – D2).5)
Where D =
2 =  - 
 is: Atan2((Px2 + Py2).5, Pz)
 is:
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25. Focusing on the ARM Manipulators in terms of Pc:
2 = Where D =
Atan2((Px2 + Py2).5, Pz) -
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26. One Further Complication Must Be Considered:
This is called the d2 offset problem
A d2 offset is a problem that states that the nth frame has a non-zero offset along the Y0 axis as observed in the solution of the T0n with all joints at home
(like the 5 dof articulating robots in our lab!)
This leads to two solutions for 1 the So-Called “Shoulder Left” and “Shoulder Right” solutions
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27. Defining the d2 Offset issue
Here: ‘The ARM’ might contain a prismatic joint (as in the Stanford Arm – discussed in text) or it might be the a2 & a3 links in an Articulating Arm as it rotates out of plane
A d2 offset means that there are two places where 1 can be placed to touch a given point (and note, when 1 is at Home, the wrist center is not on the X0 axis!)
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28. Lets look at this Device “From the Top” – a ‘plan’ view of the structure projected to the X0 Y0 plane F1 
11
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29. Solving For 1: We will have a Choice of two poses for 1: ME 3230
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30. These lead to two 2’s (Spherical)
This device (like the S110 robots in lab) is called a “Hard Arm” Solution
We have two 1’s
These lead to two 2’s (Spherical)
One for Shoulder Right & one for Shoulder Left
Or four 2’s and 3’s in the Articulating Arm:
Shoulder Right  Elbow Up & Down
Shoulder Left  Elbow Up & Down
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31. Now, lets look at Orientation IKS’s
Orientation IKS again relies on separation of joint effects
We (now) know the first 3 joints control positions
they would have been solved by using the appropriate set of equations developed above
The last three (wrist joints) will control the achievement of our desired Orientation
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32. Note: target orientation is a ‘given’ for the IKS model!
These Ideas Lead to an Orientation Model for a Device that is Given by:
This ‘model’ separates Arm Joint and Wrist Joint Contribution to the desired Target Orientation
Note: target orientation is a ‘given’ for the IKS model!
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33. Focusing on Orientation Issues
Lets begin by considering ‘Euler Angles’ (they are a model that is almost identical to a full functioning Spherical Wrist as defined using the D-H algorithm!):
Step 1, Form a Product:
Rz1*Ry2*Rz3
This product becomes R36 in the model on the previous slide
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34. Euler Wrist Simplified:
this matrix, which contains the joint control angles, is then set equal to a ‘U matrix’ prepared by multiplying the inverse of the ARM joint orientation sub-matrices and the Desired (given) target orientation sub-matrix:
NOTE: R03 is Manipulator dependent! (Inverse required a Transpose)
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35. Continuing: to define the U-Matrix
Rij are terms from the product of the first 3 Ai’s (rotational sub-matrix)
ni, oi & ai are from given target orientation
– we develop our models in a general way to allow computation of specific angles for specific cases
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36. Now Set: LHS (euler angles) = RHS (U-Matrix)
U as defined on the previous slide! (a function of arm-joint POSE and Desired End-Orientation)
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37. With C we “know” S = (1 - C2).5 Hence:  = Atan2(U33, (1-U332).5)
Solving for Individual Orientation Angles (1st solve for ‘’ the middle one):
Selecting (3,3) C = U33
With C we “know” S = (1 - C2).5
Hence:  = Atan2(U33, (1-U332).5)
NOTE: leads to 2 solutions for !
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38. Re-examining the Matrices:
To solve for : Select terms: (1,3) & (2,3)
CS = U13
SS = U23
Dividing the 2nd by the 1st: S /C = U23/U13
Tan() = U23/U13
 = Atan2(U13, U23)
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39. Continuing our Solution:
To solve for : Select terms: (3,1) & (3,2)
-SC = U31
SS = U32 (and dividing this by previous)
Tan() = U32/-U31 (note sign migrates with term!)
 = Atan2(-U31, U32)
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40. Summarizing:  = Atan2(U33, (1-U332).5)  = Atan2(U13, U23)
Uij’s as defined on the earlier slide!
– and –
U is manipulator and desired orientation dependent
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41. Let consider a Spherical Wrist:
Same Origin Point
Here drawn in ‘Good Kinematic Home’ – for attachment to an Articulating Arm
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42. IKSing the Spherical Wrist
Frames
Link
Var  d a 
3  4 4 R 4
-90
4  5 5 5
+90
5  6 6 6 d6
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43. Writing The Solution: Uij’s as defined on the earlier slide! – and –
U is manipulator and desired orientation dependent
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44. Let’s Solve (here I try the ‘Pure Inverse’ Technique):
Note: R4’s Inverse
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45. Simplifying
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46. Solving: Let’s select Term(3,3) on both sides:
0 = C4U23 – S4U13
S4U13 = C4U23
Tan(4) = S4/C4 = U23/U13
4 = Atan2(U13, U23)
With the givens and back-substituted values (from the arm joints) we have a value for 4 and the RHS is completely known!
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47. Solving for 5 & 6 For 5: Select (1,3) & (2,3) terms C5 = U33
S5 = C4U13 + S4U23
C5 = U33
Tan(5) = S5/C5 = (C4U13 + S4U23)/U33
5 = Atan2(U33, C4U13 + S4U23)
For 6: Select (3,1) & (3, 2) terms
S6 = C4U21 – S4U11
C6 = C4U22 – S4U12
Tan(6) = S6/C6 = ([C4U21 – S4U11]/[C4U22 – S4U12])
6 = Atan2 ([C4U22 – S4U12], [C4U21 – S4U11])
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48. Using the Pure Inversing:
We removed ambiguity from the solution
We were able to solve for joints “In Order”
Without noting – we see the obvious relationship between Spherical wrist and Euler Orientation!
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49. Summarizing: 4 = Atan2(U13, U23) 5 = Atan2(U33, C4U13 + S4U23)
6 = Atan2 ([C4U22 – S4U12], [C4U21 – S4U11])
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50. Lets Try One: Cylindrical Robot w/ Spherical Wrist
Given would be a Target matrix from Robot Mapping! (it’s an IKS after all!)
The d3 “constant” is 400mm; the d6 offset (the ‘Hand Span’) is 150 mm.
1 = Atan2((dx – ax*150),(dy-ay*150))
d2 = (dz – az*150)
d3 = ((dx – ax*150)2+p(dy-ay*150)2)
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51. The Frame Skeleton:
Note “Dummy” Frame to account for Orientation problem with Spherical Wrist – might not be needed if we had set wrist kinematic home for Cylindrical Machine!
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52. Solving for U:
NOTE: We needed a “Dummy Frame” to account for the Orientation issue at the end of the Arm (as drawn) – this becomes a “part” of the arm space – not the wrist space!
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53. Simplifying:
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54. Subbing Uij’s Into Spherical Wrist Joint Models:
4 = Atan2(U13, U23) = Atan2((C1ax + S1az), ay)
5 = Atan2(U33, C4U13 + S4U23) = Atan2{ (S1ax-C1az) , [C4(C1ax+S1az) + S4*ay]}
6 = Atan2 ([C4U21 - S4U11], [C4U22 - S4U12]) = Atan2{[C4*ny - S4(C1nx+S1nz)], [C4*oy - S4(C1ox+S1oz)]}
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