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by: Immanuel Brakus PhD
Immanuel Brakus PhD
GPA 3.66


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Class Notes
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This 219 page Class Notes was uploaded by Immanuel Brakus PhD on Monday October 19, 2015. The Class Notes belongs to ECSE 4490 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 235 views. For similar materials see /class/224784/ecse-4490-rensselaer-polytechnic-institute in ELECTRICAL AND COMPUTER ENGINEERING at Rensselaer Polytechnic Institute.



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Date Created: 10/19/15
23 ComputedTorque Control Topics Be able to 39 use the LagrangeEuler dynamic model equations to control a robot represent the manipulator dynamic model as a set of rstorder differential equations design a computedtorque controller design a proportional plus derivative gravity controller and explain its advantages Fall 20w Manipulator State Equation Models Reca11Dq639j CCM39 hq 194 T For an n link manipulator this is a set of n secondorder nonlinear differential equations Note D 1q exists Let XT qT VT where v q39 then can be expressed as a set of first order differential equations ComputedTorque Control Strategy Feedback a signal that cancels the effects of gravity friction the manipulator inertia tensor and Coriolis and centrifugal forces This cancellation will be done by using the Lagrange Euler dynamic model Note All of these effects were treated as disturbances in the single j 0th controller Let D 15 AD estimate of D an error To cancel the nonlinear effects consider the feedback law I What does the control law look like on a block Flaw 716 Computedtony controller Interpretatzon h 5 I attempt to cancel the nonlinearities How computed torque control com If the estimates are exact what are the closed loop dynamics If you choose K and L diagonal Example Implement the computedtorque control law for the twoaxis planar robot Solution The equations of motion of the arm are 2 2 2 quot 1 z mzaz mlala1C2 2 71 qz IImZ al mzala2C2 3 q 2 T 42 m2alazsz qlq27 30 ml mz lzcn 7mZJaIC1 2 bl41 39 2 I mzalaZCZ mzaz q 2 2 3 1 mazz39j maaS392 gmaC 232 2 2121qu o 22212b2q2 Let e r q then the control law is ComputedTorque Control summary 0 accounts for the dynamic interactions among the links unlike using a control loop around each joint 0 try to cancel thenonlinear terms due to gravity inertia Coriolis and centrifugal forces Example Find a computed torque control law for the one axis robot with a load at the tip m L Solution PDGraVity Controller Recall v v D qT cq v 419 bv The gravity controller block diagram is Figure 723 PDgrlvily manual From the block diagram e r q r KeL hq The39state equations become q v 1 D391qKe L Cq vbV Advantages The control law requires no knowledge of Dq cq v or bv Example Implement a PDGravity Controller for the two axis planar robot Solution Using the equations of motion from the computedtorque control law example identify the gravity loading terms Let e r q and find expressions for 171 and T2 Example A threeaxis SCARA robot is shown below Po c l I m xquot quot I I vl 39lx c v39 C x The dynamic model for the third axis neglecting friction is 13 2 quot7393 quot307773 a Write the state equations for the dynamic model of joint 3 Example continued b A proportional plus derivative PD gravity control law for joint 3 is given by 133 k3e3 l3e393 g0m3 where e3 r3 q3 With this controller nd the characteristic equation of the closedloop system c Find the gains in the controller so that the closedloop system will have a damping ratio of 7 z nnr 4 chiral Fran11nnn r I39 1 4 Coordinate Transformations Part I Topics Be able to de ne the direct kinematic problem use dot products nd the transformation matrix that relates one coordinate system to another 0 nd the inverse coordinate transformation carry out a rotation transformation Ref Schilling pp 18 35 Fall 2004 Goal Find the relationship between the joint variables and the tool position Tool 5quotquot new with GD Shuuldlr Too m u Figur 218 Alpha II mbolic um am Fiunn 1 Link uuuminulcs uI hu Alpha II rubolic arm Coordinate Transformations Coordinate frames will be attached to the links of a robot link 0 link n frame L k Coordinate Transformation Matrix T source destination e g Ttool I base Direct Kinematics In order to control the location of the tool we need to know the relationship between the joint variables and the actual position of the tool Dot Product x0y special case x TX 2 2 length or Euclidean norm of x which leads to the following properties of the dot product In general x y refers to the quotamountquot of xin the direction of y or Coordinate Frames De nition Coordinates p Note p projection of p onto each x k Coordinates of p in two dimensions 1315 Figure 24 Coordinates of p with re spect to the coordinate frame X Coordinate Transformations Goal Develop a sequence of transformations from tool to wrist wrist to elbow elbow to shoulder back to the base Example Singleaxis Robot Find the coordinates of p wrt F f 1 f f given the coordinates of p wrt M Note W pF Hevolute joint Figure 25 Singleaxis robot Proposition LetAlg fquot m1 1 s kj s n then Proof pJF A MM Start with the coordinates of p wrt fk Consider k 1 P 2 A11 19 A12 mg A13 17 01 Exan pleSingleaxis Robot Suppose pM 06 05 14T Find the coordinates Ofp wrt Am pf0 06 14T Solution PF A MM Inverse Coordinate Transformations Assuming F and M are orthonormal and A Me FthenhowdoesF gt M Furthermore Example Use the previous example to find pM from I9F Solution Since NF A 191 Rotations We need rotations and translations to specify the position and orientation of the end effector Fundamental Rotations Rotate M about one of the unit vectors in F Example Rotate the frame by cl about the f axis I z 5 Find the resulting coordinate transformation 1 o o matrlx R1 AnsR1 o msd ish 0 saw my Solution example cont 0 1 o quotBindiomm Repeat fgr a rotation about f M RM new 0 m 0 and for ajrotation about f 3 A RL m 1M 0 531W c0545 0 0 o 1 I I Rotation Example p 35 Suppose pM 2 0 3T and p 7r3 about the f1 axis Find the coordinates of p in the xed coordinate frame F Ans pF2 2598 1517 Solution Now suppose qF 3 4 0T Findqw M Ans q 3 2 3464T Solution 20 Single Joint Controller Design Topics Be able to 0 nd the Damping Ratio C and the Natural Frequency 0quot for the Single Joint ClosedLoop System 0 explain the limitations of this Classical Control Approach determine and explain the meaning of the bounds on K6 and K 0 nd the steadystate errors for a step and ramp input to the controller Ref 39 quotMotion Control quot A Desrochers Encyclopedia of Robotics Wiley 1988 Schilling 74 7 5 pp 256276 Fan 2004 Fi n B Pui an mnmller with vnlodty feedback quotke A 39 2202 eat KLIS Jw war14 WKe I fl Damping Ratio and the Natural Frequency nK0 KI con 21 R Je 2 MnKe K RJe Note C depends on Consider the range on J eff which depends on Joint Minimum VallgNo Load Maximum ValueNo Lo39ad Maximum ValuciFull Load number 0 kgm1 kg39m 14 7 6176 9570 3590 6950 1300 GMMN o 5 m c 1 a t Limitation of the Classical Control Approach Consider joint 1 and let C 2 Co J ie all parameters are constant except Je then for joint 1 Conclusion 0 must choose K6 and K for the worst case classical control solution may be inadequate Determination of K9 and K Design requirements overshoot is not tolerable con not near er the structural resonant frequency The Structural Resonant Frequency Let ke be the effective stiffness of the robot joint so that 8 em Je e 23 m structural resonant frequency cont What should the Bode plot of the closedloop system look like 650w 9001 Choose on so that or frequencies are suppressed Choosing on x or could excite the structural modes of the system causing vibrations Bounds on K and K 1 wquot s car or 2 nKe K s 1 kg 27 J R4 2 Jeff Wm Innbl39 h 39 HS HiSI l 5 4 151317 7 5 6 37699 3 T 20 756616 4 01 IS 1 77 5 L IS 942677 6 00 10 1256636 to determine keg and so K9 3 29 note we kwJo w k U Bound on K C 2 1 avoids an underdamped controller from 22 RB KIKb K19 Z c e 2nKe K RJe 1 0139 R861 1909 Hg 2 2mm6 K RJe Steady State Errors 0 a measure of the ability of a system to follow a step input ramp input or acceleration input step input ramp input Recall the general case ES Rrs O gtCs Find Es Use the nal value theorem to nd the steady State error ML M in mm quot0 ham Position and Velocity Errors for a Single Joint Recall 19 65o Es 72K6 K GS ssRJeff R8917 KIKb KKI steady state position error Steady state velocity error RB K KK Au em v39nllf39 9 B I pm mamaaaldsgq mn uv 05 l 15 2 25 3 35 4 Time s 39 Figure 1 Ertor due to a ramp input PM immaculde mniuv l 15 25 3 Time s Finn 3 Error due to trapezoidal input quot 35 Other Design Problems There are disturbances acting on the system fIn friction torque of motor and tach 1x torque produced by the load rs gravitational torque 3 1x If minim 13 Trajectory Planning Part 11 Continuous Path Motion Topics Be able to explain the difference between path planning and trajectory planning incorporate a desired speed pro le into the con guration vector for straight line motions 39 smooth a trajectory using interpolation techniques including parabolic blending Fall21704 Path and Trajectory Planning Path denotes the chronological history of the position and orientation of the manipulator Trajectory Trajectory Planning may be performed in Cartesian or joint space Trajectory Planning Algorithms should relieve the user from specifying every single point Note planned trajectories are generated onlin Trajectory Planning Issues All joints should complete their motion at the same time Maximum velocities and accelerations must be taken into account Typically done in Cartesian space by specifying velocity and acceleration pro les Consider yo 010 00 x 0 100 gt Tool Trajectory Recall the tool con guration vector wq 1901 i expqnnr3qT Let I be a path in the tool con guration Space Incorporate speed along the path by specifying s39t e g 3 0 I What does st represent We need the tool con guration vector that stays on this path and satis es the speed requirement along the path Example Determine the joint velocities for the Rhino robot to follow the trajectory wt a13 z40cz l a2 d51 tT00 1T for 0 g t s T Solution 0 1 0 3 a3 a4 1 o o s 0 note from p 65 rg wome 0 o 1 5 11 a2 d5 0 0 0 1 ltT requires the tool to follow a vertical path ZO Algorithm 341 Inverse Kinematics of a FiveAxis Articulated Robot Rhino XR3 q1 atan2 w2 w1 q atan2 C1w4 S1w5 W5 b1 C1w1 81w2 aAC234 d5S234 b2 d1 quot 345234 150284 W3 IIbIIZaEa q3 iarccos 2a2a3 qu atan2 a2 a3C3b2 a383b1 a2 213C3b1 a33b21 q5 1r In wg w wg n Straight Line Motion Generalize the previous result Let W0 2 initial point w1 final point Find wt as a function of st to move in a straight line over T seconds n 1 Interpolated Motion Usually only knot points points along the path and endpoints are specified The trajectory planner must produce a smooth trajectory gt Suppose we have a sequence of m knot points in tool configuration space that the tool should go through Wt and WU must be continuous why Cubic Polynomial Paths Find the cubic polynomial path wt at3 th ctd Osts T to make a robot move from w0 to w1 duiing 0 T starting With velocity v0 and ending with velocity V Solution Use the four constraints to find a b c d Armdzwn 6 ATv1v 72wlrw hrrTv 2v 3wlw T3 T2 Linear Interpolation with Parabolic Blends 510 This path isn39t smooth and requires 0 acceleration at the knot points i mquot wwwmmwgmwmlsxsa Yool commune Now the path is piecewiselinear Use parabolic blends to smooth adjacent line segments quot quot Tnol moldinle Huntll HomwheMmuinurpoh oawilhlpunmhkndJsks Traverse Wk39l to wk in time Tk with velocity Vk then parabolic blends cont Now compare v2 with v1 and suppose they are unequal For a finite acceleration at T1 change the velocity during T1 ATT1 AT with a constant acceleration note at T1 AT the tool is at the same location as it would have been on the linear segment Since the acceleration is constant the tool trajectory over T1 AT T1 AT is Mt Find a b and c assume AT is known Wt Apply the constraints wwaD 2 1 wU ADE JL T2 also wT1 AT Summary The complete blended trajectory is O 3 ts T1 AT Tl ATstsT1AT T1ATstsT1T2 Video Examples Visual Servoing Dynamic Hand Eye Coordination Robot VisualServo Control High Performance Visual Servoing Visual Head Tracking and Slaving for Visual Telepresence 3 Basic Robot Engineering Problems Topics Using a Mobile Robot Environment explain 0 task planning trajectory planninggenerating the inverse kinematics problem 39 robot control 0 the robot kinematics problem 0 joint space coordinates and world coor mates Also explain the above for an Nlink robot A Mobile Robot Environment Serves as a 2D demonstration of the various robot engineering problems Steering is achieved by adjusting 6 Need a coordinate system on the robot to x the meaning of 0 The environment contains obstacles and a coordinate system is needed to x the location of the objects or obstacles Accomplishing a Task with a Mobile Robot y Get to B while v avoiding D the obstacles A B x Basie Robot Engineering Problems Task Planning basic robot problems cont Trajectory Planning The trajectory is in world coordinates How does this translate into 9t Trajectory Generation Motion Control Accomplishing a Task With an N link Robot Task 1 Hoducnd cu m Mk 139 mlyzlr a l I wk Marian typ Tniocmrv rm Rabat rm Arm Mannr mnwlllf dynamic 3 4 4 5 7 8 Figure 91 T Fm l Tf j sum 1 Tool r 59 These subproblems are common to all robot systems 1kaj Reduced data Figure 9 cont rt motion type qt T0 Wt Task Planner Inputs Task speci ed by the user Sensory information Output Gross Motion Path Utilizes Mobile Robot Example Using a speci c algorithm Schilling p 375 you can nd the shortest collision free path E Full194 11 mm m yum Trajectory Planner Input Gross Motion Path Output Joint Angles Utilizes Interpolation Technique The Inverse Kinematics Problem This will lead to a description of the problem in ioint space coordinates trajectory planner cont Mobile Robot Example Get to B while avoiding the obstacles Robot Controller Input desired joint coordinates motion type actual joint angles Output individual joint torques Utilizes Mobile Robot Example Note outnut of the controller is a motor Arm Dynamics This is the actual arm 0 Note torques are not the actual actuating signal Could insert the equations of motion and you would have a simulation Output Mobile Robot Example note we are only concerned with steering at this point Arm Kinematics Input joint angles Output location of the tool in Utilizes The Arm Kinematics Problem Mobile Robot Example Where is the cart in the world coordinate system Trajectory Generator What about the translational motion of the cart How fast should it go Let st be the length of the trajectory starting om A then A typical speed pro le is m If there is a quotcornerquot at A1 and A2 then use Cart Location in the World Coordinate System From 6a measurement calculate Summary Figure 91 p 358 is a generic block diagram relevant to all robotic systems The rest of the course will focus on 6 Link Coordinates Topics Be able to 0 de ne and explain the four kinematic parameters associated With a robot assign orthonormal coordinate frames to an n joint arm using the Denavit Hartenberg representation Full 20114 F l G U R E I 5b PUMA Programmable universal Manipulator for Assembly manip ulator Goal 1 assign coordinate frames to each robot link 2 nd the general arm equation describing the kinematic motion of the links nu H KINEMATIC PARAMETERS FO THE ALPHA I IDIOT Mk I l l a Ilal I In 50 m o 1 1 I 0 I711 null on g 3 2 g I77l Ill 0 39 0 5 I IZSJ ml 0 10 1 Kinematic Parameters 2 joint parameters 2 link parameters u all LNV yu a xi Note the 1 Link axis is assigned x k391 Attach coordinate frames 2quot 391 is along the joint axis def 6k joint angle def dk joint distance Note for a revolute joint amp for a prismatic joint Link Parameters Between Joints 0 due to mechanical design k Note x is a common normal def 21k link length def ock twist angle Assigning Coordinate Frames xed base link 0 tool link n Figure 217 In Fig 2 17 the origin of the tool frame is the tool tip Fig 217 is the convention for the rest of this book The Denavit Hartenberg Representation a systematic notation for assigning orthonormal coordinate frames to the links Let Lk be the frame associated with link k General Approach Assign coordinate frames Add kinematic parameters Example Microbot Step 0 Number the joints from 1 to n starting with the base and ending with the tool yaw pitch and roll Step 1 Assign a coordinate frame L0 to the robot base Step 2 Align zkwith the axis ofjoint k 1 Step 3 39 Locate the origin of Lk at the intersection of z k and z k 1 When there is no intersection use the intersection of z k with a common normal between 2 k and z kquot 1 Step 4 Select x k to be orthogonal to 2k and zk 1 lfz k and z k 1 are parallel point x k away from k 7 1 z Step 5 39 Select ykto form a right handed orthonormal coordinate frame Step 7 Set the origin of LH at the tool tip Align z11 with the approach vector y11 with the sliding vector x11 with the normal vector Shoqu Elbow Taol pitch Result 757 11 X 1 x 2 1 x x V I IV x4 25 Y Tool Tao lo roll Y 55 Ful ll WemepmHmbodcn Kinematic Parameters TABLE M KINEMATC PARAMETERS FOR THE ALPHA I W Am I d a a Hm I In 2110 mm 0 39I2 0 1 O 0 l77l mm 0 3 O 0 71l quotIII B 0 3 o o I ll n Aquot n n n Step 8 Locate point bk at the intersection of the Xk and zquot391 axes Step 9 6k is the angle of rotation from X 1 to x k measured about 2 1 ie 6k is determined from the righthand rule Step 10 dk distance from origin of frame Lk1 to bk measured along 2quot Step 11 ak distance from bk to origin of frame Lk measured along xk 31 32 a3 a4 a5 Step 12 oak angle of rotation from 21 1 to 2k about xk 0 1 0 2 0 3 0 4 The Arm Matrix Want to transform from coordinate frame k to kl Strategy successively translate and rotate frame k l to make it coincident with coordinate frame k Recall steps 912 of the DH algorithm step 9 step 10 step 11 step 12 Find the composite homogeneous transformation matrix to do this Video Examples Assembly 1 Stanford University Dual Arm Assembly with the Stanford Arm 2 Adept Hard Disk Drive Assembly 3 CMU Transistor Insertion and Wire Harness Assembly 4 NASA Space Station Strut Assembly 1 Introduction to Robot Dynamics and Control Topics Be able to o classify robots based on their geometry 0 describe the basic components of a robot 0 describe two types of robot motion explain the difference between world and joint coordinates Ref Chapter I Schilling pp 1 8 Fall 2004 Robot Definition Schilling A robot is a softwarecontrollable mechanical device that uses sensors to guide one or more ende ectors through programmed motions in a workspace in order to manipulate physical objects Robot Institute of America A robot is a re programmable multi functional manipulator designed to move material parts tools or specialized devices through variable programmed motions for the performance of a variety of tasks Robot Shapes and Sizes FIGURE 12 FIGURE Lm FIGURE Lab 3 E R U m F FIGURE LA F G U R E 1 5b PUMA Programmable universal manipulator for Assembly manip uhrar FIGURE 151 Basic Components Joints and Links and an ende ector Work Envelope Conclusion You need six degrees of freedom 6 DOFs to specify the position and orientation of an object in three dimensional space There is more than one way to specify a point in three dimensions eg Maj or Axes What would a robot look like that implemented the Cartesian approach to specify a point in 3D What would a robot look like that located its endeffector using cylindrical coordinates Major axes cont Spherical coordinates TABLE 12 ROBOT WORK ENVELOPES BASED ON MAJOR AXES Abbot Axis I Axis 2 Axis 3 Total mint Camsian P P P 0 Cylindrical 39 R P P 1 Sphmcal R R P z SCARA R R P 2 Articulated R R R 3 Features Direct drive cnnslmclion AdepL ulnlruller and sniiware Iligh speed and lnw rear cycle times Large work volume High repeatability Handles large end eiTeciurs Bene ts High reliability low maintenance and mi perrurmance degradatinn nVL r lime Complete wnrktull iunLrul and fully integrated notions Higher wcrkceil lhrmlghpnl in actual applications Suitable for perrurining multiple tasks in one Vm kcell consistent highqualii production More cl39i39chivo i39nbuL uliliznllrm AdeptOne Robot Product Description The AdeptOne is a highperformance direct drive SCARA robot Designed for ntly performs hundreds of mechanical assembly electronic assembly and material handling tasks The Adeptoiie39s advantage is its unique direct drive construction a mi in a machine that can perform a W 9 Variety of tasks very cost elfectively e l envelope and longv t ul pk suitable rur yeri39onning multiple tasks in e 17 inch diameter hollow quill o r e The AdeptOne f aimes a arge wurk er ic str 9 prowling a clean passage for pneumatic and electrical linest Adepl s puweriiil conlruller and sollware a low users to maximize robot and workcell perfnrmance Adept s control capabilities include iuiiy Lian suc as visiun guidance and owe sensing which are not available on most other robots An Adeptoiic rtlbtli provides a powerml re iable asis for high ouality production Westinghouse A full line of clean room robots designed for semiconductor fabrication UNIMATE PUMA Clean Room Robots UNIMATE PUMA clean room robots are specifically designed to meet the rigorous demands of semiconductor fabrication All robots in the series are Class 10 clean room compatible fully electric and include six axes of motion for maximum dexterity The robots feature VALTM communication protocol PUMA clean room robots are based on the performance proven structure of the original PUMA industrial robots the most widely used articulated arm design in the world in addition to features specific to each model all PUMA clean room robots in corporate sealed arm link struc tures and specially prepared exterior surfaces to insure Class 10 environment compatibility Robot Material Handling Improves Productivity As chip densities increase and feature geometries decrease wafer fabrication processes become more sensitive to conv tamination Material handling with clean room PUMA robots mini mizes contamination and im proves productivity Increased die yield from fewer contamination defects Improved line yield through re duced wafer breakage Predictable throughput from more consistent operation More efficient throughput opti mizes use of capital equipment Improved worker safety by reduced exposure to hazard ous materials Superior Dexterity and Motion Control The PUMA sixaxis revolute arm provides a high degree of dexter ity for intricate and delicate handling of IC wafers VAL ll allows complex moves in tool 4 and world coordinates as well as straight line motions to be pro grammed quickly and easily using simple English commands Full Line For All Applications UNIMATION supplies a full line of clean room robot models With this wide range of size and performance features there is a PUMA clean room robot ideal for nearly every task in the semicon ductor fabrication processone supplier for all clean room robot needs PUMA 2600 Compact High Precision The PUMA 2600Fl is the small est in the PUMA clean room robot line it is ideally suited for transferring wafers between cas settes and boats inspection and precise placement on critical process equipment fixtures Its compact size facilitates incor poration into process equipment PUMA 2600 loading graphite boat AEG Westinghouse Industrial Automation Corporation XL Con guration XR3 Series Brass bushings and nylon sleeves for Standard hand r 39 long life in all joints Mlcraswiich for hard home on each axis Open construction so students can easily learn and observe Mode or rugged vr inch aircraft grade aluminum Large aluminum gears for maintenance free Operation posillonal feedback 39 sincaders an all motors provide Optlcaliy encoded DC servo motors Rerriovable encoder covers Hollow waist drive shaft The Rhino XR3 Series robotic rm provides tremendou sophistication and versatility that is unequaled b an er instructional robotic system Thequot xis arm lus gri er is an articulated tabletop robot that Is widely u or education training and research it is easy 0 use durable and lunclions learning The open construction of the arm helpst e user 9 r i e ca 0 e a any industrial application A full ine 39 39esea h c very tea ore of the arm is designed to emulate industrial technology while facilitating permits an infinite range of configurations ROBOTS mic r l y will llr yr 7 tihit minr4 i x 9 The quotteach controlquot only from UMlMicrobot The TeachMover is the only lowcost robotic arm with the features offered by an industriallevel teach con trol UMIMicrobot s teach control the arm to execute a series of movements the operator simply positions the arm using the v rious keys on the teach control and then records that position in memory by pressing the record button The operator repeats t 395 process for each desired position As many as 53 positions may be recorded in the unit s memory With the nished program in memory it can be veri ed stepiby step edited and run any number of times The teach control gives the TeachMover treme ou tenhal For example it can exhibit lowlevel intelligence by preprogramming the gripper an then sensingits status The TeachMover can also be programmed to pause for as many as 255 seconds or until it receives an external signal to move at 10 different speeds and to perform numerous other functions in the simulation of fullasize industrial robots MiniMover5 for microcomputer interface The TeachMover s precursor the MiniMoverS features the same arm construction but has no teach control Instead it interfaces directly to Apple 11quot and Tits 80 computers UM Microbot39s proprietary ARMBASIC commands written in extended BASIC language are speci cally designed to operate the MiniMover S and greatly simplify the mechanics of pro rning for individuals who are primarily interested in highly complex programs Although perhaps not as convenient as the TeachMover the price of the MiniMover S is signi cantly lower Shoulder Joint UPPE39Ar39 z ants El Major Structural Components Drive Systems Electric motors Chain drives Hydraulic motors Direct Drive Motion Classi cation Pointtopoint What are some pointtopoint applications Continuouspath following Examples of continuouspath following applications World Coordinates and Joint Coordinates Sketch of atypical world coordinate system Sketch of atypical joint coordinate system eg major axes of the PUMA World Coordinates and Joint Coordinates cont In which coordinate system is the gripper location specified Where are the control variables specified world or joint coordinates How do you locate the gripper using the control variables 25 Impedance Control Topics Be able to explain the impedance control concept nd the impedance control law for a robot manipulator Fall 2004 Motivation Control of joint angles e g the computed torque method is not the only control problem Problem How do you control the tool tip when it is in contact with the environment One approach to this problem is impedance control Goal Control the tool tip to maintain contact with the surface Strategy Mechanical Impedance Concept Consider the tool quotpressingquot against the environment The resulting deformation creates an external F Recall We can model the surface as a spring Analysis Induced joint torques and forces p 182 Work done by the tool is dW Work done by the joints is dW Controller Design Objective Generate a speci c endof arm stiffness modeled by Note Choose K to control the relationship between force and displacement For position control along dimension i For force or moment control Impedance Controller Leteru Consider the following control law Analysis of the Impedance Controller What are the generalized forces acting on the arm Dq5139 6014 Mg 176 ClosedLoop Impedance Controller X qT VT Where v c or q39 v then using q Inserting the control law Equilibrium Points For 26t fx tt the 336 that satisfy 0 f3 et are called equilibrium points What are the equilibrium points of the closed loop impedance controller 1 if k ltlt hi little deformation of the environment then 2 if k gtgt h1 tool is stiffer than the environment then Example Find an impedance control law for the one axis robot that has a load mL at the tip Assume that K and L are diagonal and u p ER 2 Solution Example Find an impedance control law for a three axis SCARA robot Note that only tool tip position can be controlled ie u p ER 3 Solution 3quot gt x T I x39 2 u z 1 v 21 y a quot 3 gt quot Use the modeling results from section 66 a1C1 a2C142 ICI a1C1a2C1 2 2 a S a S 1 1 2 1 2 Cl 151 Cz a2S172 C3 615 f 1 1 7 la C12 S1 2 0 a1C1a2C1 2 51 2 C1 2 0 a151a251 2 0 1 d1 q3 0 0 0 1 T0301 Find the manipulator J acobian Find the gravity loading Find the impedance feedback control law Example Find an impedance control law for a fouraxis cylindrical robot The arm matrix from tool to base is S1 4 C1 4 0 q3C1 Tm1q C14 S1 4 0 q3S1 base 0 0 1 qzd4 O 0 0 1 Solution What is the form of the control law What are Find the manipulator Jacobian First we need S1 C1 0 13Cl CI S1 0 1351 0 0 1 12 0 0 O 1 T03q Find the gravity loading 8 Inverse Kinematics Topics Be able to de ne the inverse kinematics problem 0 de ne tool con guration space and joint space 0 nd the tool con guration vector 0 nd the task con guration vector compute the inverse kinematics of a threeaxis planar robot Fall 2004 The Inverse Kinematics Problem Problem Statement Given a desired position p and orientation R for the tool solve tool R01 5 pq for base 0 0 o a 1 We want the transformation of Cartesian or workspace coordinates into joint coordinates Motivation necessary for realtime control of robot manipulators The desired motion of the tool is speci ed in the workspace Find the corresponding joint angle motion Example Specify a homogeneous transformation matrix that will position the peg just above the hole ie find Note that the peg extends 2 inches beyond the jaws of the gripper Solution have loll 20 I 30 you an Con guration of the tool refers to Direct kinematics problem Inverse kinematics problem kinematics equations Inverse kinematics equations Joint space Toolconfiguration In General Properties of Solutions A closedform solution to the inverse kinematics problem exists only for certain classes of robot mechanisms The solution is not unique How do multiple solutions arise Kinematic Redundancy Multiple solutions may even exist with a Nonredundanf Robot Tool Con guration Vector speci es tool con guration Without using 19 R Tool Orientation Recall R r1 r2 r3 Strategy attach roll angle information to the approach vector by sealing the approach vector by some function of the roll angle note 13 is a unit vector try fqn Task Configuration Vector Def Let p and R denote the position and orientation of the tool frame relative to the base frame and let qI1 represent the roll angle then task con guration tool con guration task con guration vector vector vector Example Show how on can be recovered from W Solution Example SCARA Robot Find the tool con guration vector for a four axis I39ObOt Ans wp 172p3 o o quot 1 Solution E El El 9 in f 9 lil E h MMhliI m From w how many DOFs does this robot have Why are there two zeros in W Inverse Kinematics of a ThreeAxis Planar Articulated Robot Derive the kinematic equations for the robot below and then solve the inverse kinematic equations Solution Find the link coordinate diagram Step 0 Number the joints from 1 to n starting with the base M l l PhiJ4 Ami h Mm Step I 39 Assign x0 yo 2 so that 2 aligns with the axis of joint 1 Step 2 Align Zk with the axis of joint k l Step 3 Since 2k and Zquot1 don39t intersect locate the origin of Lk at the intersection of zk with a normal between 2k and 2 Step 4 Since 2k and 21quot1 are in parallel point Xk away from Z Step 5 Select yk to make a right handed coordinate system Step 6 done for all joints Step 7 Tool tip is L3 Align Z3 with the approach vector y3 with the sliding vector Steps 813 Use the gures below The result is gure 31 1 example cont The Arm Matrix tool C6t an 0c SmkSer akCBk l 2 3 39 C CO S CO e T T0 T1T2use7 H k b O Suk Cock dk ase 0 0 O I l example cont The Arm Matrix Final result C S123 0 a2C12a1C1 123 S C 0 51212 alS1 m 2 123 123 b 0 o 1 d3 0 o 0 1 The tool con guration vector is W W1 W2T p is obtained from the last column of the arm matrix w K example cont Inverse Kinematics 112 Shoulder Joint Am 411ws1w5za distance from L0 to L2 depends only on q2 example cont Inverse Kinematics Base J 0 im AMW401mczwa2sma45 sz we know 12 solve for ql now W1 a1C1 aZCIZ W2 3131 32512 i base joint cont Singularity ifa1 a2 a then C1 and S1 Now if q2 1 C1 and S1 otherwise 611 Def xgt0 tan 1yx atan2yx 160 sgny7t2 xlt0 tan 1yx 3571101 1 atan2yx tan 1yx while keeping track of What quadrant y and x are in example cont Inverse Kinematics Tool Roll Joint A11vq3139tlnws W6 6Xpq3 so 613 This solution assumes that w is given perhaps from a trajectory planner or knowledge about the task l Video Examples Rovers l In atable Rover Demonstration Jet Propulsion Lab Pasadena CA 2 A Robotic Reconnaissance and Surveillance Team University of Minnesota 3 An Innovative Space Rover with Extended Climbing Abilities Swiss Federal Institute of Technology Lausanne 14 The Tool Con guration J acobian Topics nd the tool con guration Jacobian matrix 0 interpret the meaning of the matrix use the tool con guration J acobian matrix to nd the jointspace singularities use this matrix in a resolvedmotion rate controller Fall 2004 Motivation Suppose we have XU WW0 where xt is the desired position and orientation obtained from a trajectory planner Find qt Options 1 Find qt from the inverse kinematics Not so easy in general 2 Consider xt dt The Tool Configuration J acobian Matrix useful for the trajectory planning problem find qt from W WCJ1 where xt is the desired trajectory in tool configuration space One approach is to solve the inverse kinematics by brute force Consider r Vqt the tool con guration Jacobian matrix Example Find the tool con guration J acobian matrix for the Adept One Solution T Recall x alcl 2012 1 V 1 Tool jag quot1S1 2512 a LDM d1 qa d4 7 6 1quot w M W 39x r 0 39 0 expq1t VClt Interpret the dependency of the tool tip location on the joint variables Example Interpret the meaning of the tool con guration J acobian matrix for the Rhino Solution 1 Vqz has the following form see section 51 1 xxxxO X Vql X X X gtlt R N O Q where x denotes a nonzero element Joint Space Singularities We have Xt Vqt 40 and so 40 Examine Vqz 1 at certain points in the joint space this matrix does not have full rank often Vqt is not 6 x 6 Def 3 is a jointspace singularity iff What ifn lt 6 Example Boundary Singularities Check the rank of Vqt for the Adept One robot Are there any singularities Aux sfo Solution aSI azS12 ale2 0 0 alCl aZClV2 a2C12 O 0 0 0 gt1 0 Vqt 0 0 0 0 O 0 0 0 0 0 0 Expq4TETt Vqt has rank lt 4 iff the first two columns become linearly dependent for some q1t q2t Can this happen When Interior Singularities Two or more joint axes become collinear and the tool con guration remains constant even though the robot moves in joint space Example Examine the Microbot Alpha II for singularities Ans See qpnext page Solution Using the arm matrix below the tool con guration vector is c cmcfslss 4163954165 Csm s C 177r8C2 177802396552M splucsiqs5 402354 65 751s 5 511778C2177363 96552 ismcs SW95 rem a 2159 177252177ssn 95scm 0 0 0 E 1 consider 3W 40 2 aql There Will be a singularity if you can force this column to be all zeros try qBq1 BZB n 3q5T 0lt lt1t2 and a3 32 then aW010 aql ResolvedMotion Rate Control Proposition 54 1 for n s 6 9W V Il 40 For n 6 40 Motion in tool configuration space is resolved into joint space components Example Find 139 t for the Adept One where ia1Sla2S1 2 alcl 2cm 0 Vltqltrgtgt O 0 O a C 0 Let Vqt 0 0 0 then VT 610 123142 0 0 412C 0 0 0 1 0 0 0 0 0 0 0 O O expq41ITI Solution OOOOamp l NOOOOO 12 Trajectory Flaming Part 1 Topics Be able to nd the arm matrices to carry out a pick and place operation nd the arm matrices that specify a nut fastening trajectory Fall 2001 Pick and Place Operations 0 for loading and unloading machines 0 pointtopoint motion control Pick Point pick T base Wurk smut Sliding vector grasps the part along two parallel surfaces LiftoffPoint How should the liftoff point be approached Lifto point 39ame lift T base V approach distance Place Point place T 2 base place R vI 6quot F39I ure 4U Tool mn gunu39an a quotIt Wark surfac pick pail Set Down Point located by T set base set down point is near the place point place Knowing T base then 39 T Set base Set down point is v units from the place point Motion Control Gross motion is at high speed N0 overshoot Example Consider a robotic workstation with parts A and B Suppose the centroid of A 6 12 2T and the centroid ofB 10 5 1T 39 k a Find the arm matrlx T pm that W111 posmon base the tool to pick up part A f ornoas ove along the long sides g f A b Find the arm matrix Tb ffe that will position the tool to place A on B aligning the long sides and the centroids M a m nuns DO IA 0001 pm that will position c Find the arm matrix Tbm the tool to pick up B from above along tl39rleolgnf sides d Find the arm matrix T172136 that will position the tool to place B on A aligning the long scidleoss and the centroids e Find the task configuration vector for the mqk described in Dart d Nutfastening Trajectorxm Apply a screw transformation to frame B x yquot 21 let T seconds to roll one revolution Recall Screw A d k Rotdgtk tranOt 139quot Here 430 Let p pitch of the bolt thread in threadsmm then Mt Let b length of the thread then the total threading time rm and Mt 2 I0 this leads to T1317 2 This represents the threading operation with respect to the bolt frame B Where is the bolt frame relative to the base frame B has origin p and from the orientation of frame B bolt T 2 base The threading operation relative to the base is T thread Tbolt Tthread t base base bolt sin2ntT cos27ttT 0 r cos21ttT sin21ctT 0 0 0 0 1 h tpT 0 0 O 1 What is the task con guration vector th e d Tbagea t Wq It is a function of time now and so the inve kinematics must be solved in real time 19 SingleJoint Controllers for Robots Topics Controller Design Problems Stanford Manipulator Modeling the GearLoad System Actuator motor Modeling 0 Velocity Feedback for Adding Damping Damping Ratio and the Natural Frequency Be able to derive the closedloop transfer function for a sin lejoint controller Find the dam ing ratlo an the natural frequency What are t e cr1teria for selecting these Ref quotMotion Control quot A Desrochers Encyclopedia at Robotics Wiley 1988 Schilling sections 7 7 2 7 4 5 Fall 2001 Design Problems Assume a pointtopoint controller Each joint actuator has a position loop The desired joint coordinates rt can be obtained from the teaching mode and used as reference commands to the Closedloop controller during the playback mode We also get rt from the inverse kinematics Control Problems Control System Basics A secondorder system has the following characteristic equation s2 2Cmns 0 O The system poles depend 011C Cgt 1 overdamped SL2 cantconiCz 1 C 1 critically damped sL2 0 n Cltl underdamped 512 C0nij0nl 2 Figure 313 P0 versus Control System Basics continued Step Response vs C I v0 pvvtcul uvtrshool T rise lime z Stanford Manipulator nun Mmmuuummrm Permanent magnet DC motors Optical encoder for position feedback Tachometer for velocity feedback Figure 6 shows position velocity feedback l39ignn a Posi nn mm with nicer nibch 65 7 K lt wmgsm mm m a I 944 MOHON CONTROL k LSI1102 39 PUMA arm MLThePUMAamwnholmetul ediagrm Modeling of Each Joint Notation Problem J mm hath 1 innr nofthnjoint xmrunthaaztuam dd Flnd the J inertia afdu mnipuhmr link 3 a damping coe dent at the actuator aide BL dambing medicine load ide generated at 1 a genede mm at ch amm shah L tum1 1nd mm the actuator 03I I angular displacement at the ammtuerha e39 angular diaphumin N 39Nnmhntpuu h nthnn ahir duham 39 Model the Actuatorgear Zoad System Figure 7 ngradnsn actuatorgear load system cont At the load side of the gears JL as on the actuator side OIT m Using the gear ratio in 1 I m 2 3 4 5 Actuator Modeling Find 0msVs Atrium VIM Combining 8 and 9 and taking the LaPlace transform Vs 10 Typical Motor Data Sheet SXXX MOTOR SIZE DATA 2539 C um H van Lam 900 92 L 5 um Lam nsxx L1 actuator modeling com write 5 in the frequency domain Tms 12 Using 10 with L 0 15 13 Substitute 13 into 12 and note that both sides are in terms of Vs and 6ms 67 14 Open Loop Transfer Function Find 65sEs Figure 5 A jam poaitiun wnmuex Actuator voltage comes from an error signal 60 then Vs hand 65s then the open loop transfer function is ems Em l8 Open Loop Transfer Function for Figure 6 Find 6SsES for figure 6 and then use it to 93 S find What is the purpose of velocity d s feedback in figure 6 In figure 5 the motor fed back In figure 6 we have eso F l Em S 9 Closing the loop Determination of K0 and K K9 and K determine the damping ratio and the natural frequency 0 overshoot is not tolerable natural frequency of the closedloop system should be away from the structural resonant frequency Problem Find an expression for the natural frequency and the damping ratio in terms of K6 and K Solution From 20 the characteristic equation is 7 The Arm Equatlon Topics Be able to de ne and use screw transformations construct the homogeneous transformation matrix that maps frame k coordinates into frame 1 coordinates 39 nd the composite coordinate transformation matrix arm matrix that maps tool coordinates into base coordinates Ref Schilling pp 5761 4951 The Arm Matrix Construct the homogeneous transformation matrix that maps frame k coordinates into frame kl coordinates Do this by successive rotations and translations of frame kI to make it coincident with frame k Summary of Operations TAMIC mum FRAMEk 1 TO FR Opaninn Descriplian 1 Rome Ibomz 2 Thadm alung byd 3 Tr umgx hyn 4 Raul MI by a These operations are atranslation or rotation Note Operations 1 amp 2 3 amp 4 together form a screw transformation along 2quotquot1 x kquot 1 k Concluswn T k IS a compos1t10n of two screw transformations Screw Transformations a linear displacement along an axis combined with an angular displacement about the same axis Screw 7 p k Screw pitch 2 Reverse the order of operations Rotation first then translation Return to the arm matrix From Table 2 3 Link Coordinate Transformation qquot391 Tk il qr link coordinate transformation com Final result CGk SOkCock SOkSock akCBk S k CakCBk C6kSock akSGk 0 Sock Coc d k k 0 0 0 1 Note To1 maps shoulder coordinates into base coordinates T12 maps T23 maps Also note one of the parameters in the matrix is always variable Inverse Link Coordinate Transformation Find T 1 First partition Tklil CBk S6kCak SGkSak akCGk R p Tk 86k CukCGk C6kSak akSOk M k1 r O Sak Cock dk 0 0 O S 1 O O 0 1 which we know has an inverse equal to RT 5 RTp CGC SBA 0 ak eCakSGk Cup cek Suk dA 501 SukSQk ASa Cek Cut dkCu 0 0 0 J Am Tiquot The Arm Matrix In general tool T q base where qk is the k th joint variable and n is the number of links Partition the problem tool 39 wrist tool T q T q T q base base wrist For a six axis robot arm equation cont tool 134 5 PM 3 base 0 o 0 s 1 Example Microbot Compute the arm matrix using the link coordinate diagram below Solution wrist T base Insert Link 7 l 5 h 32 17 Parameters x 4 I 3 from Table 22 g3 la LT quot a w 976 E quot 1 TABLE 24 KNEMATC PARAMETERS FOR THE ALPHA ll ROBOT Axis 9 d a a Home 39 x a 2150 nun 1r2 o solutlon cont 2 a 0 mm 039 u 3 0 0 1773 mm 0 0 4 9n U 0 1r1 1rZ s a 1295 mm o o o F162 is rsi nzC Cz c3 753 0 3c SIC 73 s CI CICZS C151 3951 CI 2C239 JCB SICZ v quot51313 CI SI 2C2 73E23 0 dfazs2 0 0 1 0 53 quotC23 0 d1 25quot 3523 U 0 0 0 0 0 l D O 0 1 251F2 5 C 0 353 Note C1 C2 C3 C1 82S3 cos 61cos zcos 63 sinezsines solution cont tool T wrist tool wrist tool TC1 T lt1 T C1 base base wrlst cszazsss c cmssslc5 csm a C117xczunxcuwsssm sycmcfclsS SCmSS CCS ssm a 31177sCZ117sCBr9sssm 234 5 234 5 734 S C S S C i 2159 177SSZ 177 3S3923 965Cm 3512 Video Examples Recon gurable Robots Recon gurable Robot the robot changes its geometry Basic idea alter the shape of the robot for a variety of tasks 1 2 and 3 RPI Tetrabot 4 Polybot 21 Positioning Errors with Single Joint Controllers Topics Be able to nd the position and velocity errors for a singlejoint control system 0 explain the sources of disturbances acting on the arm specify control gains to reject disturbances o nd the steadystate errors when there are disturbances acting on the arm design a control system to compensate for steadystate errors Ref 39 quotMotion Control quot A Desrochers Encyclopedia of Robotics Wiley 1988 pp 94395 Schilling 7 4 75 pp 256276 Fall 2001 Figure 8 Position unbuller with valodky feedblck e g 7 K9 I 20 s eau gqupwqu 4 ne I Position and Velocity Errors for a Single Joint Recall 19 es s nKe K I G s Es ssRJe RBe KIKb KKt SteadyState position error e lim S1S p sao 1 Gs Steady state velocity error e lim S1S2 ssv PO 1 How can you make em gt 0 Other Design Problems There are disturbances acting on the system fIn friction torque of the motor and tach 1X torque produced by the load H 17g gravitational torque a In 7 T m Conclusion Need to consider these when calculating the steady state errors Disturbances on the Arm Reevaluate the steady state errors in View of the disturbances in Figure 9 Need new Es What is the torque required to move the load now 32 The feedback signals generate the motor command so that Tm s 33 Equate 32 and 33 and solve for 95s torque required to drive the load torque produced by torque to overcome the disturbances the motor Recall Ds F ms Tgs nTxs nKI K69 13 nRF ms Tgs nTxs 34 RJe s 2 RBe KIKb KKts nKeKl 938 If all disturbances 0 then 34 20 In 34 Gs S has a component due to 0d S and another due to the disturbances Ds Example Disturbance Rejection In Figure 9 combine all disturbances into 9s S d T S and evaluate the T ds and nd H as DC gain How does this in uence the choice of K9 Solution SteadyState Position Error Need E s 9ds 053 Let Qs RJe s 2 RBe KIKb K1915 quotK9 K1 Assume that the disturbances are constant Cf C Cx Fms Tgs g ms s s 3 Find the steady state error when 6d 3 9 s EIT 0139 Compensation Strategy If you know or can estimate fm 1g TX then feed these forward What should Tas and Tds be to minimize essp 11 T m l v1 r ya 4 m I E t mull mekhum hm Use the analysis of Figure 9 and replace Fms nTxs Tgs with 3 9 eSSp can be made zero by choosing TaS Tds Conveyor Following Velocity Error 940 I I V mm is updated to synchronize With the moving conveyor 6ds Cvs 2 em lim sEs 5 0 velocity error cont The robot is required to be overdamped so recall RBe KIKb KKI 2 2nKBKIRJe 2 30 so e 2 44 SW To stay away from the structural resonant frequency on s or2 which led to 29 J 2 R K6 3 4nK I 29 and so 44 becomes gt em Velocity Error Compensation Feed forward another term Vds Find Vds mm ma 4 x K VJ 23 xxx x A quot311 an ms 23 mggl 1 I x Figu lL Fodtnrwud ammunition for dw ImdyIHII vulodty cant From the new steadystate velocity error is Eliminate the first term by choosing Vd s Conclusion Need feedforward compensation to reduce errors to zero 2 Introduction to Robotic Systems Topics Be able to 0 describe the sensors that are used in robotic systems 0 interpret robot speci cations explain the difference between repeatability precision and accuracy of robotic mechanisms explain the meaning of servocontrolled robots Full 2w Sensors Vision Force Torque Position Cross re Others Load Effects As soon as the object is picked up the moment of inertia of each manipulator link changes The control system may have to change its gains to handle the disturbance of the load Robotic Systems Robots working with sensors and other robots Camera Ian 1 robot bus Example For the system below given camera tool mouse base 2 Tbasel TbaseZ 3 Tbasel 7 and Tbasel camera tool H 39 base 1 mouse base 2 camera a Fmd an expressmn for Tmom b Find 0 Tmouse I Solution Robot Speci cations see product speci cations attached Speci cations load carrying capacity repeatability maximum speed coordinate system maximum movements drive system motion control system Loaa Carrying Capacity weight of the tool part to be carried Inertia is also important since 2 grippers or loads may weigh the same but have a different inertia Adept ne Robot Product Speci cations ylnad Downward Iorce Joint 4 lnenin sum Br I maximum 20 lb 40 lb 100 hltln2 000 lbin Positioning cu Remlution Repeats 1y Accuracy Options Highvaocurncy Positioning System BPS Force Sznning Module Clean room packng Class 10 Variabl Cnmplimoe Wrist VCW lL Uur Connection 5 E n 2 n n 1 a 3 User signal lines and Calihrltion Method Um locnl calibration indies Cycle 39l lme Maximum velocity no land The what tool performs a continuouspath motion from location a back a quotaquot The path cansilm af all straightline segment as she 30 RAM Paylond Cycle m l 1 09 m 13 lbs 13 m 20 lbs 17 we r mm pouxbk mu m num mnmm and mmu39rua Induazd me n we a 17 x Tuna mu Menu Hunmum Fauin anina 511m raps 25 mbob m wnlmller cable set System Requirements Controller Compatibln with Adept Aseries or Sverieu MC and CC conuuuen power The Adept eomroller mull all aw power md contra so psi mimum 110 psi maximum 1 cm clean dry as axtluding and gum mmmenu Environment 41 1 39F s 50 C 5 90 relative nnmldlry nonwndenning AdeptOne Dimensions A lnner link B Outer Iin Maximum reach Ki 7 Minimum reach 150 Ruse Orchard Way San Jose CA 95134 Phone 408 4320888 Telex 171942 Telefax 408 4328707 Florianstraase 3 34500 Dortmund 1 West emany Phone 0231129081 Telex 8227463 Telelax 11231125089 For the AdeplOne robot or other Adept products contaci your local sale engineer or call the Adept Inlor motion Hotline at l 8002923878 Cdquot odepv r lechnoloav inc hl39 Specifications MMi Sonia GENERAL all models Ftevoluta Axes External Program Storage Double SidedDouble Denslty v Electric DC Servos Floppy Disk System Com uter Floor 0r overhead Controller Mounting Teaching Method Teach Pendant andor computer Gripper Control 4vway pneumatic solanmd terminal interconnect Cable Length 164 5m Program Language VAL ll MODEL 250 562 761 762 Repeatablmv 20002 in 005mml 0004 in 01mm 0008 In 02mm 0005 in 02mm Payload 22 lb 10kg 55 lb 25kg 220 lb 100kg 440 lb zunkg 53 lb 40m 39 Reach To cenler at erSi 163 in 414mm 341 in 866mm 590 In 1500mm 490 in 1250mm To iool flange 102 ln i462mm 362 m 920mm 63910 In 1625mm 540 in 1372mm Program Capacity 19kW lSkW 17kW 17kW CMOS User Memory 40 inIsecl max 10msec max 40 lnsec max 10msec max 490 insec max 192 lnsec max 123msec max Dimsec max up to 3232 3232 3232 32132 Straight Line Velocity IO Capacity 1276 lb 5800kg 208240450 VAC ZSOOW 1298 lb 5900kg 208240480 VAC 2500W 29 lb 132ng 110130 VAC 1200W 40 It 30kg 11012081240 VAC 1500W Arm Wol hi Power Requirements Ambient Operating 50120 F 1050 C 50420 F lo50 C 50120 F lo50 C 50120 F lo500 Temperature loam relative humidity noncomansmg enigma a lmol lnouslnal llho llucluailons no human electrostall discharge 39 Larger payload available wilh console controlllr Controller 250552 Size 125 In H x 175 in W x x 196 In D Flack Mounl 318mmH x 445mmW x 499mmD Analog IO Module Weight 80 lb 368w Additional quot0 Capacity 562761762 Size 457 in H x 235 in W x 315 in D IBM PC supervisor software Console 1160mmH x SDOmMW x BOOmmD industrial Versions Available 400 lb 182kg UNIVAL Robot Control lF3954NEMA 12 Options and Accessories UNIViSlONquotM Vision System Weight Environment Work Envelopes NC For dunleg lmldiinirull drowmg emml uNlMAIION canierilne Side View Top View swam 11 umnng mm mm mm hltp lwwwslaublicomquotpenDOCumenIKLLanguagFGB Robotics mum GB Product Range m0 RXQQ RX1 Q RX17Q RX90 Number of axis 455 Nominal payload 6 k Maximum payload 12 kg each 98050rgm Repeatability t rnrn Control Unlt gf CS7M Programming language Work envelope RX90 Mm RX90 L 0025 mm 087 057M V Specifications UW M 74 3 Axis Cartesian Drives with up to a 2 revolute axis wrist Teach Pendant Manual Data input Oitlene Programming Contiguratlon Teaching Methods Axis Feedback AC Servomoters X amp YAxis Rack amp Pinion ZAxis Ball Screw Closed Loop Resolver MANIPULATOR UNIMATE 6000 UNIMATE 6000 XL XAxis Travel 1050 it 10 it Y Axls Travel 46 it 1050 l L ZAxis Travel 3 it 3 ft Wrist AAxis 1160 t160 BAxis i lOS 1105 Velocity X ii YAxas 1800 lPM 1800 IPM ZAxis 1200 IPM 1200 IPM 20 RPM 20 RPM Wrist Axes Repeatability per Linear Axis 3200 in 1005 In Accuracy per Linear Axls 005 in par it not to exceed 1008 in lull travel oi axes Payload Capacity 4 0 lbs vertical 6000 lbs on YltAxis Tabla Manipulator Weight 9 100 lbs 000 lbs 39Irnproved Aocuracy and Repeatability is available as an option CONTROLLER CNC UNIVAL Programming Language RS274D VAL II 4 0 Architecture IBM PCAT VME Bus Offline Teach and Repeat 240 480 VAC 60 Hz 3 Phase 96quotH x 58quotW x 32D Programming Method Power Requirements Controller Dimensions Air Conditioning UNIMA TE 6000 Configuration 33 4000 BTU 2000 Teach and Repeat 240 480 VAC 60 Hz 3 Phase 9639H x 58quotW x 31 D BTU UNIMATE 6000 XL Con guration Specifications XR3 SERIES ROBOTIG ARM With either leach pendant or computer control the Xi robotic arm is attractive to students at all levels whethsr computer ltiiterate or advanced programmers It can be used 0 teec ro c movement strategies collision avoidance robot salary and many other subjects that require opportunities tor hundwn experience Because at Its sophistication llexlblllty and usefulness as an tor instructional purposes by Junior and senior high schools voca ndiiltn il cation programs and graduate schools as well as motor manu locturing companies and government laboratories such as General Motors and NASA FEATURES VERSATlLE The 5 axis arm plus gripper together with It 8 axis controller has many aitferent applications tor teaching Industrial training and research These nclude leaching principles otheuaie nonce Introducing s en 0 the funda mentals at tnoustriai automation developing workceils teaching programming techniques conducting leaslbllltv studies learning mec onle or rebel operation and teaching robotic move strateg es TRULV EMUIATES INDUSTRiAL ROBOT Like Industrial robots and all Xi Series equipment the Xi robotic arm uses servo motors These are rr re bot VALUE The Xi Series robotic arm costs a traction at the price at c be equipped with XR robots and selected accessories tor less than the single unit price 01 most industrial robe PROVIDES HANDSON EXPERtENCE The XR Series robotic arm puts 9 beginning student allowing the rst day or sfu OPEN CONSTRUCTiON Like all other XR Series products the XR robotic arm has open design 0 help the user observe and under stand the mechanics 0 robot movement The external Optlccl r troubleshooting procedures OVABLE COVERS FOR OPYICAL ENCODERS The DC serve motors came with removable covers tor the optical encoders Th en on to protect the encoders E V RA The XR robotic arm can be controlled with the Mark ill teach pendant or with a host computer using Rhine s RoooiatkTM or RhinoVAL sottwore A multitude of programming Dosslaiiltles exist iO CAPABiUTiES The IIO capabilities or the controlling Mark III controller aflerd the Rhino user with emendous possibilities for workcell development NONiNTIMlDATlNG r The small size and user friendly software at r bots Use of the XR Series robotic arm eliminates the expense dlf cuiw and cancer 0 using a heavy industrial robot to train stu dents and amp oyees i GED 39 weight The Rhine Xi robotic arm is Incredibly lorgivlng to students since It is rugged and is relatively inexpensive to repair COMPREHENSIVE OWNER39S MANUAL A complete manual is included to give detailed instruction on setup operation one pro gramming ot the to i Rhino has representatives throughout the USA Canada and a large number oi other countries Please call or write for the name of the representative in your area no Gmelgn IIIInoII UH I20 9 Item manus R O B O i S ailmmmmolownc mm not Inc we mm siev ltreet Rhine Robots Inc reserves the right to change any and all specifications and prices without prior notice ENGINEERED AND MANUFACTURED IN THE USA 380009 o in robotics research for ten years In Research 39 individuals he founded UMlMicrobot in early 1980 and the product was introduced with great ford This is UMlMicrobot Y Institute Along with three other experienced management and support personnel 39 Miclvbotlnr m WORKING VOLUME Product Specifications TeachMover MiniMoverS R may mImIIL Iddlh orul puwu lupply Ind mmyutulo wlre Ipediy types Con guri nn Ftve revnlull me And lmegnl hand Five molule Ixa n testl lund r r n r cuwuu RAM tamed In hue olunlt inmee 39 mwtpullkl lululatex bind me u MILh szctAlzle belweenll 150 sou sou mo me 4300 Ind sew baud yam ngugn ASle Ihrnuyl min pun AMASICO Ipedfy interface Teldl Carlin 3 mod 13 mnc mr keyboud NA Emmi lo Power qu umenb 12 to u voltmds mp DC um 12 no H van A mp DC 0b holubl MERIan lnkdltzubllkng rmmm Fly 1 4549mm lfu muwlm munHull Remlu nn 035mm ux on nth I 05mm max on ud I l Gdde rune isN mix MN m Rem mm mm Static and Fm N m N m Ann Weight 41 Pml oning Amt 75mm sped 017M Iankl mm m muvunentgluum n tadi mntmL ee mumJ or myu p Central Mt d PIT Palm m Plant Pl Point to Point Pmymmlng Clpldly 53 stored petitions 5w yulnh depends on hm cm Hue Malian x W Slumldu Malian 1443 4539 414439 415 Elbnvl Motion r 44939 r ur wmmou em wrlumzh 90 Gripper Opening 75mm D75mm Power Sn 1 Included 105175 VAC in u VDC on I 5 un 2211 ud C Ln u l pp VAC m mm p t m tpu p ngm l v p l JBVDCaul yuLsunps mvnc nude Ctrlying an Oyuami Apple II39 Flumum NA Optional lncludu liltull baud with ARMBASICW in ROM Ind openlinn inmucuonn 112530quot Sohwire NA Optionl Lndudel ARMEASICO an mite Ind opmlimi Model L Level II huh diam Head unite us mace Anlhilhn om Dirtrlbnmn UMl Group Limited UM House U39MlMicmbat me 1233 E Mailan um Australia Pty m ushii Belgium Denmark Finland 915 51 lame Road Surbilon ViewAlviso Rnad Sunnyvale 13 Bryan Avenue Mnsmln Park rm Germany HallantL India Surrey m 4cm England Callfamia CA 94089 Penn Weslem Au min 6012 Italy Norway Sweden SwimAland Teleyhonl am me 734 um rphnne 09 us me Splin Telepnone m 39y 5m Famimile 01 399 ragmun am we 734 0107 Tel Flumlle 09 386 557 robot speci cations cont Repeatability ability of a manipulator to reposition itself This is essential information Measurements shduld be made with the same payload velocity acceleration direction of approach and ambient temperature Different regions of the workspace Accuracy different than repeatability ability of a manipulator to approach an arbitrary point in space E nonfat 1 Annuity gun llo Adjnccnl tool positions robot specifications cont What are the Sources of Repeatability Problems and Inaccuracies Maximum Tip Speed no loaa 0 It will be different when there is a load Which joint or joints were moving when this measurement was made 0 Was it straight line motion robot speci cations cont Coordinate System Maximum Movements Manipulator Reach Horizontal Reach Horizontal Stroke Vertical Reach Vertical Stroke Hullmill rudl Holkoaul muquot m Ranch am wax u yumnw robu Precision for Different Robots Consider minimum motion possible along each axis of a robot Cartesian Cylindrical Table 19 Horizontal and Vertical Precision Robot Horizontal Vertical Tvue Precision Precision Cartesian Uniform Uniform Cylindrical Decreases radially Uniform Spherical Decreases radially Decreases radially ries niform Articulated Varies Varies precision cant Cylindrical worst case What is the angular precision What is the overall horizontal precision What is the total overall precision Example SCARA The vertical precision is uniform which makes it Servocontrolled Robots Each joint axis has a closed loop The robot is programmed in the teach mode The manipulator is moved through a sequence of desired points while recording the joint angles along the way Teach Pendant Pointto point Servocontrolled Robot Typical Operating Sequence Playback Mode Position of the manipulator joints is obtained and sent to the program Next desired joint positions are sent from a single control computer Error signals are generated which actuate the joint motors When the error signals are all 0 the joints stop moving The next taught point is sent out Continuous Path Servo controlled Robots Teach Method 1 Operator runs the tool over the desired path while position andor velocity information is sampled 2 Playback is independent of the teaching speed 10 Kinematics and Inverse Kinematics of the Rhino XR3 Topics Be able to nd the kinematics and inverse kinematics of a ve axis articulated robot 0 nd the toolcon guration vector for this case Ref Schilling pp 6268 9096 M The Rhino XR3 has ve revolute joints Identify the joints from the gure below I I The Link Coordinate Diagram Link Parameters Axis 6 d a a Home UlPUJN 9 s satisfy the right hand rule d s are translations along 2 3 1 5 are translations along x s 0139s are rotations about x Elbow Tuol pitch in 91 a 4 z3 I X X The Arm Matrlx cek Caksek SukSGk akCSk TlOOl Twrist TIOOl k 59k Cakcek 5ukcek aksek T4 base base wrzst H 0 Sq Cak d k k 0 0 0 I Find T W base Example Where is the wrist frame in base coordinates when q 0 90 90 0 90T What is the meaning of this q Solution with This is the orientation and position of L3 in base coordinates Interpretation ce mags SuSQ ace armmatrzxcont k k k Tk 56k CakCGk SukC6k akS39ek H o Suk Cuk dk 0 0 0 l T1001 a wrist Final Arm Matrix Clczzacs 7 5155 7CIC13455 SICS 7C13234 5 ClaC2 7 3C2 39 4cm 7 153234 SICZCMCS 7 C155 751C545 7 C1C5 7515234 3 5a2C2 aaczs acm 7 355234 75234Cs 311455 7cm 3 dl 7 252 7 3523 7 quot45234 7 dsczaa Example What is the position and orientation of the tool in base coordinates when q 0 90 90 O 90T Solution T532 q Where is the tool tip located in base coordinates Inverse Kinematics Find the T 001 Con guration Vector Here the approach vector r3 is along 25 What is it in base coordinates This leads to the tool con guration vector W01 Base Joint Note W2w1 Elbow Joint q3 q234 q2 q3 q4 Use C1w451W5 define to find q1 substitute for the W39s 2 2 Now use b1 172 Elbow up solution kee workspace Joint limil down solution Shoulder Joint q2 we know q3 so solve b2a282a3823 for C2 and 82 Tool Pitch Joint q4 1234 12 13 14 Tool Roll Joint Angle q5 15


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