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ME 4135 Differential Motion and the Robot Jacobian Slide Series 6 Fall 2011 R. R. Lindeke, Ph.D.

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Lets develop the differential Operator – bringing calculus to Robots The Differential Operator is a way to account for “Tiny Motions” ( T) It can be used to study movement of the End Frame over a short time interval ( t) It is a way to track and explain motion for different points of view

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Considering motion: We can define a General Rotation of a vector K: By a general matrix defined as:

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These Rotation are given as: But lets remember, for our purposes that this angle is very small (a ‘tiny rotation’) If the angle is small we can have some simplifications: Cos small 1 Sin small small

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Substituting the Small angle Approximation: Similarly for Y and Z:

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Simplifying the Rotation Matricies (product) Here, we have neglected products of the terms!

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What about Small (general) Translation? We define it as a matrix: General Tiny Motion is then (both Rot and Translation):

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So let’s use this idea: Here we define a motion which is due to a robot’s joint(s) moving during a small time interval: T+ T = {Rot(K,d )*Trans(dx,dy,dz)}T Note, T is the original pose Substituting for the matrices:

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Solving for the differential motion effect ( T)

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Factoring the T out on the R.H.S.:

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Further Simplifying: We will call this matrix the del operator:

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Thus, the Change in POSE ( T) dT (that is T) = T = {[Trans(dx,dy,dz)*Rot(K,d )] – I} Here we see that this operator is analagous to the derivative operator d( )/du when taken wrt HTM’s

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Lets look into an application: Given: Subject it to 2 simultaneous movements: – –Along X 0 (dx) a translation of.0002 units (/unit time) – –About Z 0 a Rotation of 0.001rad (/unit time)

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Graphically: R Here: R init = (3 2 + 5 2 ).5 = 5.831 units init = Atan2(3,5) = 1.0304 rad

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Where is the Frame ‘n’ after one time step? Considering Position: Considering Position: –Effect of “Translation”: X=3.0002 and Y = 5.000 New R f = (3.0002 2 + 5.0 2 ).5 = 5.83105 u –Effect of Rotation fin = 1.0304 + 0.001 = 1.0314 rad –X f = Cos( fin ) * R f = 2.99505 –Y f = Sin( fin ) * R f = 5.00309

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Where is the Frame n after one time step? Considering Orientation:

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After 1 time step, Exact Position is:

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Lets Approximate it with the ‘ operator’ T new = T init + dT = T init + T init Where:

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Therefore T new is Approximately:

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Comparing: “Exact”: “Exact”: Approximate: Approximate: Realistically these are all but equal and using the ‘del’ approximation is soooo much easier!

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We can (might!) use the ‘del’ approach to move a robot in space: Take a starting POSE (T orig ) & a starting motion set (deltas in rotation and translation as function of unit times) Form operator for motion Compute dT ( T orig ) Form T new = T orig + dT Repeat as time moves forward over the time steps …

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Taking Motion WRT other Spaces (non inertial basis) Original Model: – –dT = T(1) Motion Taken WRT another (non-inertia) space: – –dT = T T (2) – –Here T implies motion wrt its own frame but could be any other non-inertia space But the change itself (dT) is independent of point of view so, equating (1) and (2) we can isolate T T = (T) -1 T

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Solving for the specific Terms in T Origin Change Vector wrt T space : Angular effects wrt T space : d, n, s & a vectors are extracts from the T Matrix d p is the translation vector in is the rotational effect in

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Subbing into a ‘del’ Form:

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An Application of this issue: T WC R T Cam Part T R part If the Part is moving along a conveyor and we “measure” its motion with a Camera (in camera frame) but we want to pick up the part with the robot, we must ‘track’ it so we need to known the part motion in robot’s space (not camera space!).

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This is a Motion “Mapping” Issue: PaRWCCaPa Pa R C Pa PaRWCCaPa Knowns: C Robot in WC Camera in WC And of course Part in Camera (But we don’t need it now!)

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Lets Isolate the “Middle” RWCCa R C RWCCa To solve for R we “mathematically move” from R to R directly ( R ) and “The long way around”:

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Rewriting into Standard Form: It can be shown for any 2 Matrices (A & B): A -1 *B = (B -1 *A) -1 (1) If, on the previous page we consider: WC T R as “A” and WC T Cam as “B”, And define ( WC T Cam ) -1 *( WC T R ) as “T” Then, Using the theorem (from matrix math) stated as (1) above (and then) Rewriting, we find that R = “T” -1 ( Cam ) ”T” R is now shown in the standard form – –the terms: d, n, s & a vectors in the “T” matrix and the d p and vectors from the Cam are then all that is required to define part motion with respect to the robot space

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R is given by simplifying ( WC T Cam ) -1 *( WC T R ) = “T” HERE: d, n, s & a vectors are extracts from the “T” Matrix above d p is the ‘translation’ vector in Cam is the ‘rotational effects’ in Cam

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Now, Lets Examine the Jacobian Fundamentally: We use these Jacobians to further our knowledge of Geometric Calculus

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In This Model, D dot & D q,dot are: We state, then, that the Jacobian is a mapping tool that relates Cartesian Velocities (of the end frame) to the movement of the individual Robot Joints

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Lets build one from ‘1 st Principles’ – Here is a Spherical Arm: R Lets start with only linear motion ---- its straight forward!

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Writing the Position Models: Z = R*Sin( ) X = R*Cos( )*Cos( ) Y = R*Cos( )*Sin( ) To find velocity differentiate these which leads to:

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Writing it as a Matrix: This is the Jacobian; It is built as the Matrix of partial joint contributions (coefficients of the velocity equations) to Velocity of the End Frame

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Here we could develop an Inverse Jacobian: It was formed by taking the partial derivatives of the IKS eqns.

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The process we did just now is limited to finding Linear Velocity We need linear and angular velocity for full functional robots! We can approach the problem by separations as before … Here we had separated Velocity (Linear from Rotational) but not joints (arms) from joints (wrist) Generally speaking, in the Jacobian we will obtain one Column for each Joint and 6 rows for full velocity effect We say the Jacobian is a 6 by n matrix

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Seperation Leads to: A Cartesian Velocity Term o V n : An Angular Velocity Term o n : Each of these “Ji’s” are 3 Row by n Columned Matrices!

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Building the Sub- Jacobians (in general): We hold 3 definitional stipulations: Velocities can only be added if they are defined in the same space Motion of the end effecter (n frame) is taken wrt the base space (0 frame) Linear Velocity effects are (truly) physically separated from angular velocity effects To address the problem we will only move one joint at a time (uses the fundamental DH separation principle!)

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Lets start with the Angular Velocity (!) Considering any joint i, its Axis of motion is: Z i-1 (Z in Frame i-1) The (modeling) effect of the joint is to drive the very next frame (frame i) If Joint i is revolute: – –here k i-1 is the unit vector of Z i-1 – –This could be applied to each of the joints (revolute) in the machine (it rotates the next frame – of course, all subsequent frames move similarly!) But if the Joint is Prismatic, it has no angular effect on its “controlled” frame and thus subsequent frames

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Developing the J We need to add up each of the joint effects BUT, We need to “normalize” them to base space to added them up DH methods allow us to do this! Since Z i-1 is in a Frame of the model, we really need only extract the 3 rd column of the product of A 1 * A 2 * …*A i-1 to have a definition of Z i-1 in base space to add the effect of Joint i in this ‘common space’ (note, the qdot term is the rotational velocity of the revolute joint)

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So the J then is just this: As stated previously, Z i-1 is the (1 st 3 rows) 3 rd col. of A 0 *…A i-1 And we will have a term in the sum for each joint Note Z i-1 for Joint i – per DH algorithm!

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For the Spherical Device: All the rest of the Z i ’s depend on the Frame Skeleton drawn! Notice: 3 cols. since we have 3 joints!

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Building the Linear Jacobian It too will depend on the movement of Z i-1 It too will depend on the movement of Z i-1 It too will require that we normalize each joints contribution to the base space It too will require that we normalize each joints contribution to the base space We will find that revolute and prismatic joints will make functionally different contributions to the solution (as if we would think otherwise!) We will find that revolute and prismatic joints will make functionally different contributions to the solution (as if we would think otherwise!) Prismatics are “Easy,” Revolutes aren’t Prismatics are “Easy,” Revolutes aren’t

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Building the Linear Jacobian

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Building the Linear Jacobian – Prismatic Joints When the prismatic joint i moves, all subsequent links move (linearly) at the same rate and in the same direction

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Building the Linear Jacobian – Prismatic Joints Therefore, for each prismatic joint of a machine, the contribution to the Jacobian (after normalizing) is: Z i-1 which is the 3 rd column of the matrix given by: A 1 * … * A i-1 This is as expected based on the model on the previous slide

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Building the Linear Jacobian – Revolute Joints This is a dicer problem – remembering the idea of prismatic joints on angular velocity … But no that won’t work here just because its a rotation, and it (primarily) changes orientation of the end, it does also have a linear contribution effect to the motion of the end – that is revolute joints have a “levering effect” which moves the origin of the n-frame (a cartesian motion). We must compute and account for this effect and then normalize it too.

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Building the Linear Jacobian – Revolute Joints Using this model we would expect a rotation i would lever the end by an effect that is equivalent to the CROSS product of the driver vector and the connector vector

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Building the Linear Jacobian – Revolute Joints This is the driver vector given by: Z i-1 X i-1 d n [with Magnitude equal to joint speed]

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Building the Linear Jacobian – Revolute Joints Z i-1 X i-1 d n is the direction of the linear motion of the revolute joint i on n-Frame motion It to must be normalized Notice: i-1 d n = 0 d n – 0 d i-1 (call this eqn 2) This “normalizes” the vector i-1 d n But the d-vectors are really origin position of the various frames (Frame i-1 and Frame n ) – ie the positions of their “Origins” So let’s rewrite eqn. 2 as: i-1 d n = O n – O i-1

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Building the Linear Jacobian – Revolute Joints The contribution to the J v due to a revolute joint is then: Z i-1 X (O n – O i-1 ) – –Where: Z i-1 is the 3 rd col. of the 0 T i-1 (A 0 *… *A i-1 ) O i-1 is 4 th col. of the 0 T i-1 (A 0 *… *A i-1 ) O n is 4 th col. Of 0 T n (the FKS!) NOTE when we pull the columns we only need the first 3 rows

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Building the Linear Jacobian Summarizing: – –The J v is a 3-row by n columned matrix – –Each column is given by joint type: Revolute Joint: Z i-1 X (O n – O i-1 ) Prismatic Joint: Z i-1

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Combining Both Halves of the Jacobian: For Revolute Joints: For Revolute Joints: For Prismatic Joints: For Prismatic Joints:

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What is the Form of the Jacobian Robot is: (PPRRRR) – a cylindrical machine with a spherical wrist: Robot is: (PPRRRR) – a cylindrical machine with a spherical wrist: Z 0 is [0,0,1) T ; O 0 = (0,0,0) T always, always, always! Z 0 is [0,0,1) T ; O 0 = (0,0,0) T always, always, always! Z i-1 ’s and O i-1 ’s are per the frame skeleton Z i-1 ’s and O i-1 ’s are per the frame skeleton

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Lets try this on the Spherical ARM we did earlier: 11 22 d3d3 The robot indicates this frame skeleton

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Lets try this on the Spherical ARM we did earlier: FrLinkVar da C S C S 0→11R 11009001C1S1 1→22R 2+90009001-S2C2 2→n3P00 d3001010 LP Table:

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Lets try this on the Spherical ARM we did earlier: A i ’s: A i ’s:

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Lets try this on the Spherical ARM we did earlier: T1 = A1 T1 = A1 T2 = A1 * A2 T2 = A1 * A2 T 0 n = T3 = A1*A2*A3 T 0 n = T3 = A1*A2*A3

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Lets try this on the Spherical ARM we did earlier: THE JACOBIAN Of This Form:

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Lets try this on the Spherical ARM we did earlier: THE JACOBIAN Here: Here:

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Lets try this on the Spherical ARM we did earlier: THE JACOBIAN

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