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ME 4135 Robotics & Control R. Lindeke, Ph. D.

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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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So this IKS is Nasty (as we know!) It a more difficult problem because: The Equation set is “Over-Specified” 12 equations in 6 unknowns Space is “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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Uses of IKS Builds Workspace Allows “Off-Line Programming” solutions Thus, compares Workspace capabilities with Programming realities to assure that execution is feasible Aids in Workplace design and operation simulations

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Performing IKS For (mostly) Industrial Robots: First lets consider the previously defined Spherical Wrist simplification All Wrist joint Z’s intersect at a point The nth Frame is offset from this intersection by a distance d n along the a vector of the desired solution (3 rd column of desired orientation sub- matrix) This result is as expected by following the DH Algorithm!

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Performing IKS We can now separate the effects of the ARM joints Joints 1 to 3 in a full function manipulator (without redundant joints) They function to position the spherical wrist at a target POSITION related to the desired target POSE … From the WRIST Joints Joints 4 to 6 in a full functioning spherical wrist Wrist Joints function as the primary tool to ORIENT the end frame as required by the desired target POSE

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Performing IKS: Focus on Positioning We will define a point (the WRIST CENTER) as: P c = [P x, P y, P z ] Here we define P c = d target - d n *a P x = d target,x - d n *a x P y = d target,y - d n *a y P z = d target,z - d n *a z

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R Manipulator – a simple and special case which can be found by doing a ‘Pure IKS’ solution R Frame Skeleton

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LP Table and A i ’sFramesLinkVarda SSSS CCCC S C 0 → 1 1R + 90 009010C1-S1 1 → 2 2P0 d2 + cl2 001010

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FKS is A1*A2:

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Forming The IKS:

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Examining these two equation (n, o, a and d are ‘givens’ in an inverse sense!!!): Term (1, 4) & (2,4) both side allow us to find an equation for as a positioning joint in the arm: (1,4): C1*(d 2 +cl 2 ) = d x (2,4): S1*(d 2 +cl 2 ) = d y Form a ratio to build Tan( ): S1/C1 = d y / d x Tan = d y /d x = Atan2(d x, d y )

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Forming The IKS: After is found, back substitute and solve for d 2 : (1,4): C1*(d 2 +cl 2 ) = d x Isolating d 2 : d 2 = [d x /C1] - cl 2

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Alternative Method – doing a pure inverse approach Form A 1 -1 then pre-multiply both side by this ‘inverse’ Leads to: A 2 = A 1 -1 *T 0 n given

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Simplify RHS means:

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Solving: Selecting and Equating (1,4) 0 = -S1*d x + C1*d y Solving: S1*d x = C1*d y Tan( ) = (S1/C1) = (d y /d x ) = Atan2(d x, d y ) Selecting and Equating (3,4) -- after back substituting solution d 2 + cl 2 = C1*d x + S1*d y d 2 = C1*d x + S1*d y - cl 2

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Performing IKS For Industrial Robots: What we saw with the -R IKS solutions will always work for a robot – with or without a spherical wrist These methods are often tedious – and require some form of “intelligent application” to find a good IKS solution Hence that is why we like the spherical wrist simplification (for industrial robots) introduced earlier – separating ARM and Wrist Effects and joints

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Focusing on the ARM Manipulators in terms of P c : Prismatic: q 1 = P z (its along Z 0 !) – cl1 q 2 = P x or P y - cl2 q 3 = P y or P x - cl3 Cylindrical: 1 = Atan2(P x, P y ) d 2 = P z – cl2 d 3 = P x *C1 – cl3 or +(P x 2 + P y 2 ).5 – cl3

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Focusing on the ARM Manipulators in terms of P c : Spherical: 1 = Atan2(Px, Py) 2 = Atan2( (P x 2 + P y 2 ).5, P z ) D3 = (P x 2 + P y 2 + P z 2 ).5 – cl3

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Focusing on the ARM Manipulators in terms of P c : Articulating: 1 = Atan2(Px, Py) 3 = Atan2(D, (1 – D 2 ).5 ) Where D = 2 = - is: Atan2((Px 2 + Py 2 ).5, P z ) is:

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One Further Complication: This can be considered the d 2 offset problem A d 2 offset is a design issue that states that the nth frame has a non-zero offset along the Y 0 axis as observed in the solution of the T 0 n with all joints at home This leads to two solutions for 1 the So-Called Shoulder Left and Shoulder Right solutions

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Defining the d 2 Offset issue Here: ‘The ARM’ might be a revolute/prismatic device as in the Stanford Arm or it might be a 2 & a 3 links in an Articulating Arm and 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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Lets look at this Device “From the Top” a Q 11 F1F1

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Solving For 1 We will have a Choice (of two) poses for :

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In this so-called “Hard Arm” We have two 1 ’s These lead to two 2 ’s (Spherical) Or to four 2 ’s and 3 ’s in the Articulating Arm Shoulder Right Elbow Up & Down Shoulder Left Elbow Up & Down

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The Orientation Model Evolves from: Separates Arm Joint and Wrist Joint Contribution to the desiredTarget (given) orientation

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Focusing on Orientation Issues Lets begin by considering Euler Angles (they are a model that is almost identical to a full functioning Spherical Wrist): Form Product: R z1 *R y2 *R z3 This becomes R 3 6

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Euler Wrist Simplified: And this matrix is equal to a U matrix prepared by multiplying the inverse of the ARM joint orientation matrices inverse and the Desired (given) target orientation NOTE: R 0 3 is Manipulator Arm dependent!

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Simplifying the RHS: (our so- called U Matrix)

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Continuing:

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Finally:

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Solving for Individual Orientation Angles (1 st ): Selecting (3,3) → C = U 33 With C we “know” S = (1-C 2 ).5 Hence: = Atan2(U 33, (1-U 33 2 ).5 NOTE: 2 solutions for !

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Re-examining the Matrices: To solve for : Select terms: (1,3) & (2,3) C S = U 13 S S = U 23 Dividing the 2 nd by the 1 st : S /C = U 23 /U 13 Tan( ) = U 23 /U 13 = Atan2(U 13, U 23 )

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Continuing our Solution: To solve for : Select terms: (3,1) & (3,2) -S C = U 31 S S = U 32 Tan( ) = U 32 /-U 31 = Atan2(-U 31, U 32 )

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Summarizing: = Atan2(U 33, (1-U 33 2 ).5 = Atan2(U 13, U 23 ) = Atan2(-U 31, U 32 )

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Let do a (true) Spherical Wrist:

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IKSing the Spherical Wrist FramesLinkVarda 3 → 4 4R 444400-90 4 → 5 5R 555500+90 5 → 6 6R 6666 d6d6d6d600

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Writing The Solution:

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Lets See By Pure Inverse Technique:

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Simplifying

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Solving: Examination here lets select (3,3) both sides: 0 = C4U 23 – S4U 13 S4U 13 = C4U 23 Tan( 4 ) = S4/C4 = U 23 /U 13 4 = Atan2(U 13, U 23 ) With the given desired orientation and values (from the arm joints) we have a value for 4 the RHS is completely known

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Solving for 5 & 6 For 5 : Select (1,3) & (2,3) terms S5 = C4U 13 + S4U 23 C5 = U 33 Tan( 5 ) = S5/C5 = (C4U 13 + S4U 23 )/U 33 5 = Atan2(U 33, C4U 13 + S4U 23 ) For 6 : Select (3,1) & (3, 2) S6 = C4U 21 – S4U 11 C6 = C4U 22 – S4U 12 Tan( 6 ) = S6/C6 = ([C4U 21 – S4U 11 ],[C4U 22 – S4U 12 ]) 6 = Atan2 ([C4U 21 – S4U 11 ], [C4U 22 – S4U 12 ])

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Summarizing: 4 = Atan2(U 13, U 23 ) 5 = Atan2(U 33, C4U 13 + S4U 23 ) 6 = Atan2 ([C4U 21 – S4U 11 ], [C4U 22 – S4U 12 ])

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Lets Try One: Cylindrical Robot w/ Spherical Wrist Given a Target matrix (it’s an IKS after all!) The d 3 “constant” is 400mm; the d 6 offset (call it the ‘Hand Span’) is 150 mm. 1 = Atan2((d x – a x *150),(d y -a y *150)) d2 = (dz – a z *150) d3 = [(d x – a x *150) 2,(d y -a y *150) 2 ].5 - 400

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The Frame Skeleton: Note “Dummy” Frame to account for Orientation problem with Spherical Wrist modeled earlier

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Solving for U: NOTE: We needed a “Dummy Frame” to account for the Orientation issue at the end of the Arm

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Simplifying:

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Subbing U ij ’s Into Spherical Wrist Joint Models: 4 = Atan2(U 13, U 23 ) = Atan2((C1a x + S1a z ), (S1a x -C1a z )) 5 = Atan2(U 33, C4U 13 + S4U 23 ) = Atan2{a y, [C4(C1a x +S1a z ) + S4 (S1a x -C1a z )]} 6 = Atan2 ([C4U 21 - S4U 11 ], [C4U 22 - S4U 12 ]) = Atan2{[C4(S1n x -C1n z ) - S4(C1n x +S1n z )], [C4(S1o x -C1o z ) - S4(C1o x +S1o z )]}

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