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Robotic Kinematics – the Inverse Kinematic Solution. ME 3230 Kinematics and Mechatronics Dr. R. Lindeke. Cartesian Space. Joint Space. Joint Space. FKS vs. IKS. In FKS we built a tool for finding end frame geometry from Given Joint data:

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robotic kinematics the inverse kinematic solution

Robotic Kinematics – the Inverse Kinematic Solution

ME 3230

Kinematics and Mechatronics

Dr. R. Lindeke

ME 3230

fks vs iks

Cartesian Space

Joint Space

Joint Space

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:

Cartesian Space

ME 3230

so this iks problem is nasty in mathematics this is considered a hard modeling issue
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

ME 3230

but the iks is very useful some uses of iks include
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

ME 3230

doing a pure iks solution the r manipulator
Doing a Pure IKS solution: the R Manipulator

R Frame Skeleton (as DH Suggest!)

Same Origin Point!

ME 3230

forming the iks
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

ME 3230

forming the iks9

If 2 Matrices are Equal then EACH and EVERY term is also Uniquely equal as well!

Forming The IKS:
  • 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)

ME 3230

forming the iks10
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

ME 3230

alternative method doing a pure inverse approach
Alternative Method – “doing a pure inverse approach”
  • Form A1-1 then pre-multiply both side by this ‘inverse’
  • Leads to: A2 = A1-1*T0ngiven

ME 3230

simplifying solving
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

ME 3230

performing iks for industrial robots a more involved problem
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!

ME 3230

performing iks with a spherical wrist
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 POSITIONrelated 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

ME 3230

performing iks focus on positioning
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)

ME 3230

treating the arm types as separate robot entities
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)

ME 3230

cantilevered cartesian robot
Cantilevered Cartesian Robot

J2

J3

X0

P-P-P Configuration

Y0

Z0

J1

ME 3230

gantry cartesian robot
Gantry Cartesian Robot

P-P-P Configuration

X0

Z0

Y0

J3

J2

J1

ME 3230

cylindrical robot
Cylindrical Robot
  • Doing IKS – given a value for:
    • X, Y and Z of End
    • Compute , Z and R

Z0

R

Y0

X0

ME 3230

spherical robot
Spherical Robot
  • Developing a IKS (model):
  • Given Xe, Ye, & Ze
  • Compute ,  & R

R

Z0

Y0

X0

ME 3230

2 link articulating arm manipulator
2-Link Articulating Arm Manipulator

Z0

3

L2

L1

X0

Y0

2

1

ME 3230

focusing on the arm manipulators in terms of p c
Focusing on the ARM Manipulatorsin 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}

ME 3230

focusing on the arm manipulators in terms of p c23
Focusing on the ARM Manipulatorsin terms of Pc:
  • Spherical:
      • 1 = Atan2(Px, Py)
      • 2 = Atan2( (Px2 + Py2).5 , Pz)
      • D3 = (Px2 + Py2 + Pz2).5 – cl3

ME 3230

focusing on the arm manipulators in terms of p c24
Focusing on the ARM Manipulatorsin terms of Pc:
  • Articulating:
      • 1 = Atan2(Px, Py)
      • 3 = Atan2(D, (1 – D2).5)
        • Where D =
      • 2 =  - 
        •  is: Atan2((Px2 + Py2).5, Pz)
        •  is:

ME 3230

focusing on the arm manipulators in terms of p c25
Focusing on the ARM Manipulators in terms of Pc:
  • 2 = Where D =

Atan2((Px2 + Py2).5, Pz) -

ME 3230

one further complication must be considered
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

ME 3230

defining the d 2 offset issue
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 1is at Home, the wrist center is not on the X0 axis!)

ME 3230

lets look at this device from the top a plan view of the structure projected to the x 0 y 0 plane
Lets look at this Device “From the Top” – a ‘plan’ view of the structure projected to the X0 Y0 plane

F1

11

ME 3230

solving for 1
Solving For 1:
  • We will have a Choice of two poses for 1:

ME 3230

this device like the s110 robots in lab is called a hard arm solution
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

ME 3230

now lets look at orientation iks s
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

ME 3230

these ideas lead to an orientation model for a device that is given by
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!

ME 3230

focusing on orientation issues
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 becomesR36 in the model on the previous slide

ME 3230

euler wrist simplified
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)

ME 3230

continuing to define the u matrix
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

ME 3230

now set lhs euler angles rhs u matrix
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)

ME 3230

solving for individual orientation angles 1 st solve for the middle one
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 !

ME 3230

re examining the matrices
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)

ME 3230

continuing our solution
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)

ME 3230

summarizing
Summarizing:
  •  = Atan2(U33, (1-U332).5)
  •  = Atan2(U13, U23)
  •  = Atan2(-U31, U32)

Uij’s as defined on the earlier slide!

– and –

U is manipulator and desired orientation dependent

ME 3230

let consider a spherical wrist
Let consider a Spherical Wrist:

Same Origin Point

Here drawn in ‘Good Kinematic Home’ – for attachment to an Articulating Arm

ME 3230

writing the solution
Writing The Solution:

Uij’s as defined on the earlier slide!

– and –

U is manipulator and desired orientation dependent

ME 3230

solving
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 andthe RHS is completely known!

ME 3230

solving for 5 6
Solving for 5 & 6
  • For 5: Select (1,3) & (2,3) terms
      • 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])

ME 3230

using the pure inversing
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!

ME 3230

summarizing49
Summarizing:
  • 4 = Atan2(U13, U23)
  • 5 = Atan2(U33, C4U13 + S4U23)
  • 6 = Atan2 ([C4U22 – S4U12], [C4U21 – S4U11])

ME 3230

lets try one
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).5 - 400

ME 3230

the frame skeleton
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!

ME 3230

solving for u
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!

ME 3230

subbing u ij s into spherical wrist joint models
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)]}

ME 3230