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**Introduction to Robotics**

4/14/2017

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**General Course Information**

The course introduces the basic concepts of robotic manipulators and autonomous systems. After a review of some fundamental mathematics the course examines the mechanics and dynamics of robot arms, mobile robots, their sensors and algorithms for controlling them. two robotic arms everything in Matlab (and some Java) 4/14/2017

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**A150 Robotic Arm link 3 link 2 Symbolic Representation of Manipulators**

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Kinematics the study of motion that ignores the forces that cause the motion “geometry of motion” interested in position, velocity, acceleration, etc. of the various links of the manipulator e.g., where is the gripper relative to the base of the manipulator? what direction is it pointing in? described using rigid transformations of the links 4/14/2017

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**Kinematics forward kinematics: inverse kinematics:**

given the link lengths and joint angles compute the position and orientation of the gripper relative to the base for a serial manipulator there is only one solution inverse kinematics: given the position (and possibly the orientation) of the gripper and the dimensions of the links, what are the joint variables? for a serial manipulator there is often more than one mathematical solution 4/14/2017

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Wheeled Mobile Robots robot can have one or more wheels that can provide steering (directional control) power (exert a force against the ground) an ideal wheel is perfectly round (perimeter 2πr) moves in the direction perpendicular to its axis 4/14/2017

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Wheel 4/14/2017

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Deviations from Ideal 4/14/2017

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Differential Drive two independently driven wheels mounted on a common axis 4/14/2017

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Forward Kinematics for a robot starting with pose [ ]T moving with velocity V(t) in a direction θ(t) : 4/14/2017

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**Sensitivity to Wheel Velocity**

σ = 0.05 σ = 0.01 4/14/2017

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**Localization using Landmarks: RoboSoccer**

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Maps goal * start 4/14/2017

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Path Finding goal start 4/14/2017

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Localization goal start 4/14/2017

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**EKF SLAM Application www.probabilistic-robotics.org**

[MIT B21, courtesy by John Leonard]

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**EKF SLAM Application raw odometry estimated trajectory**

[courtesy by John Leonard]

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**Introduction to manipulator kinematics**

Day 02 Introduction to manipulator kinematics 4/14/2017

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**Robotic Manipulators a robotic manipulator is a kinematic chain**

i.e. an assembly of pairs of rigid bodies that can move respect to one another via a mechanical constraint the rigid bodies are called links the mechanical constraints are called joints Symbolic Representation of Manipulators 4/14/2017

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**A150 Robotic Arm link 3 link 2 Symbolic Representation of Manipulators**

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**Joints most manipulator joints are one of two types**

revolute (or rotary) like a hinge allows relative rotation about a fixed axis between two links axis of rotation is the z axis by convention prismatic (or linear) like a piston allows relative translation along a fixed axis between two links axis of translation is the z axis by convention our convention: joint i connects link i – 1 to link i when joint i is actuated, link i moves Symbolic Representation of Manipulators 4/14/2017

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Joint Variables revolute and prismatic joints are one degree of freedom (DOF) joints; thus, they can be described using a single numeric value called a joint variable qi : joint variable for joint i revolute qi = qi : angle of rotation of link i relative to link i – 1 prismatic qi = di : displacement of link i relative to link i – 1 Symbolic Representation of Manipulators 4/14/2017

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**Revolute Joint Variable**

qi = qi : angle of rotation of link i relative to link i – 1 link i qi link i – 1 Symbolic Representation of Manipulators 4/14/2017

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**Prismatic Joint Variable**

qi = di : displacement of link i relative to link i – 1 link i – 1 link i di Symbolic Representation of Manipulators 4/14/2017

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**Common Manipulator Arrangments**

most industrial manipulators have six or fewer joints the first three joints are the arm the remaining joints are the wrist it is common to describe such manipulators using the joints of the arm R: revolute joint P: prismatic joint Common Manipulator Arrangements 4/14/2017

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**Articulated Manipulator**

RRR (first three joints are all revolute) joint axes z0 : waist z1 : shoulder (perpendicular to z0) z2 : elbow (parallel to z1) z0 z1 z2 q2 q3 shoulder forearm elbow q1 waist Common Manipulator Arrangements 4/14/2017

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**Spherical Manipulator**

RRP Stanford arm z0 z1 d3 q2 shoulder z2 q1 waist Common Manipulator Arrangements 4/14/2017

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SCARA Manipulator RRP Selective Compliant Articulated Robot for Assembly z1 z2 z0 q2 d3 q1 Common Manipulator Arrangements 4/14/2017

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Forward Kinematics given the joint variables and dimensions of the links what is the position and orientation of the end effector? a2 q2 a1 q1 Forward Kinematics 4/14/2017

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**Forward Kinematics choose the base coordinate frame of the robot**

we want (x, y) to be expressed in this frame (x, y) ? a2 q2 y0 a1 q1 x0 Forward Kinematics 4/14/2017

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Forward Kinematics notice that link 1 moves in a circle centered on the base frame origin (x, y) ? a2 q2 y0 a1 q1 ( a1 cos q1 , a1 sin q1 ) x0 Forward Kinematics 4/14/2017

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Forward Kinematics choose a coordinate frame with origin located on joint 2 with the same orientation as the base frame (x, y) ? y1 a2 q2 y0 q1 a1 x1 q1 ( a1 cos q1 , a1 sin q1 ) x0 Forward Kinematics 4/14/2017

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Forward Kinematics notice that link 2 moves in a circle centered on frame 1 (x, y) ? y1 a2 ( a2 cos (q1 + q2), a2 sin (q1 + q2) ) q2 y0 q1 a1 x1 q1 ( a1 cos q1 , a1 sin q1 ) x0 Forward Kinematics 4/14/2017

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Forward Kinematics because the base frame and frame 1 have the same orientation, we can sum the coordinates to find the position of the end effector in the base frame (a1 cos q1 + a2 cos (q1 + q2), a1 sin q1 + a2 sin (q1 + q2) ) y1 a2 ( a2 cos (q1 + q2), a2 sin (q1 + q2) ) q2 y0 q1 a1 x1 q1 ( a1 cos q1 , a1 sin q1 ) x0 Forward Kinematics 4/14/2017

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Forward Kinematics we also want the orientation of frame 2 with respect to the base frame x2 and y2 expressed in terms of x0 and y0 y2 x2 a2 q2 y0 q1 a1 q1 x0 Forward Kinematics 4/14/2017

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**Forward Kinematics without proof I claim: x2 = (cos (q1 + q2),**

sin (q1 + q2) ) y2 x2 y2 = (-sin (q1 + q2), cos (q1 + q2) ) a2 q2 y0 q1 a1 q1 x0 Forward Kinematics 4/14/2017

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Inverse Kinematics given the position (and possibly the orientation) of the end effector, and the dimensions of the links, what are the joint variables? y2 x2 (x, y) a2 q2 ? y0 a1 q1 ? x0 Inverse Kinematics 4/14/2017

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Inverse Kinematics harder than forward kinematics because there is often more than one possible solution (x, y) a2 y0 a1 x0 Inverse Kinematics 4/14/2017

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**Inverse Kinematics law of cosines (x, y) a2 b q2 ? y0 a1 x0**

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**Inverse Kinematics and we have the trigonometric identity therefore,**

We could take the inverse cosine, but this gives only one of the two solutions. Inverse Kinematics 4/14/2017

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**Inverse Kinematics Instead, use the two trigonometric identities:**

to obtain which yields both solutions for q2 . In many programming languages you would use the four quadrant inverse tangent function atan2 c2 = (x*x + y*y – a1*a1 – a2*a2) / (2*a1*a2); s2 = sqrt(1 – c2*c2); theta21 = atan2(s2, c2); theta22 = atan2(-s2, c2); Inverse Kinematics 4/14/2017

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**Inverse Kinematics Exercise for the student: show that**

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Day 03 Spatial Descriptions 4/14/2017

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**Points and Vectors point : a location in space**

vector : magnitude (length) and direction between two points 4/14/2017

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Coordinate Frames choosing a frame (a point and two perpendicular vectors of unit length) allows us to assign coordinates 4/14/2017

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Coordinate Frames the coordinates change depending on the choice of frame 4/14/2017

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Dot Product the dot product of two vectors 4/14/2017

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Translation suppose we are given o1 expressed in {0} 4/14/2017

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Translation 1 the location of {1} expressed in {0} 4/14/2017

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Translation 1 the translation vector can be interpreted as the location of frame {j} expressed in frame {i} 4/14/2017

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**Translation 2 p1 expressed in {0} a point expressed in frame {1}**

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Translation 2 the translation vector can be interpreted as a coordinate transformation of a point from frame {j} to frame {i} 4/14/2017

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Translation 3 q0 expressed in {0} 4/14/2017

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Translation 3 the translation vector can be interpreted as an operator that takes a point and moves it to a new point in the same frame 4/14/2017

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**Rotation suppose that frame {1} is rotated relative to frame {0}**

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Rotation 1 the orientation of frame {1} expressed in {0} 4/14/2017

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Rotation 1 the rotation matrix can be interpreted as the orientation of frame {j} expressed in frame {i} 4/14/2017

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Rotation 2 p1 expressed in {0} 4/14/2017

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Rotation 2 the rotation matrix can be interpreted as a coordinate transformation of a point from frame {j} to frame {i} 4/14/2017

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Rotation 3 q0 expressed in {0} 4/14/2017

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Rotation 3 the rotation matrix can be interpreted as an operator that takes a point and moves it to a new point in the same frame 4/14/2017

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**Properties of Rotation Matrices**

RT = R-1 the columns of R are mutually orthogonal each column of R is a unit vector det R = 1 (the determinant is equal to 1) 4/14/2017

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**Rotation and Translation**

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Rotations in 3D 4/14/2017

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Day 04 Rotations 4/14/2017

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**Properties of Rotation Matrices**

RT = R-1 the columns of R are mutually orthogonal each column of R is a unit vector det R = 1 (the determinant is equal to 1) 4/14/2017

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Rotations in 3D 4/14/2017

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Rotation About z-axis +'ve rotation 4/14/2017

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Rotation About x-axis +'ve rotation 4/14/2017

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Rotation About y-axis +'ve rotation 4/14/2017

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**Relative Orientation Example**

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**Successive Rotations in Moving Frames**

Suppose you perform a rotation in frame {0} to obtain {1}. Then you perform a rotation in frame {1} to obtain {2}. What is the orientation of {2} relative to {0}? 4/14/2017

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**Successive Rotations in a Fixed Frame**

Suppose you perform a rotation in frame {0} to obtain {1}. Then you rotate {1} in frame {0} to obtain {2}. What is the orientation of {2} relative to {0}? 4/14/2017

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**Composition of Rotations**

Given a fixed frame {0} and a current frame {1} and if {2} is obtained by a rotation R in the current frame {1} then use postmulitplication to obtain: Given a fixed frame {0} and a frame {1} and if {2} is obtained by a rotation R in the fixed frame {0} then use premultiplication to obtain: 4/14/2017

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**Rotation About a Unit Axis**

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**Rigid Body Transformations**

Day 05 Rigid Body Transformations 4/14/2017

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**Homogeneous Representation**

translation represented by a vector d vector addition rotation represented by a matrix R matrix-matrix and matrix-vector multiplication convenient to have a uniform representation of translation and rotation obviously vector addition will not work for rotation can we use matrix multiplication to represent translation? 4/14/2017

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**Homogeneous Representation**

consider moving a point p by a translation vector d not possible as matrix-vector multiplication always leaves the origin unchanged 4/14/2017

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**Homogeneous Representation**

consider an augmented vector ph and an augmented matrix D 4/14/2017

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**Homogeneous Representation**

the augmented form of a rotation matrix R3x3 4/14/2017

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**Rigid Body Transformations in 3D**

{1} {0} {2} 4/14/2017

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**Rigid Body Transformations in 3D**

suppose {1} is a rotated and translated relative to {0} what is the pose (the orientation and position) of {1} expressed in {0} ? {1} {0} 4/14/2017

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**Rigid Body Transformations in 3D**

suppose we use the moving frame interpretation (postmultiply transformation matrices) translate in {0} to get {0’} and then rotate in {0’} to get {1} {0’} {0} {0’} {1} {0} Step 1 Step 2 4/14/2017

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**Rigid Body Transformations in 3D**

suppose we use the fixed frame interpretation (premultiply transformation matrices) rotate in {0} to get {0’} and then translate in {0} in to get {1} {0} {0’} {1} {0} Step 1 {0’} Step 2 4/14/2017

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**Rigid Body Transformations in 3D**

both interpretations yield the same transformation 4/14/2017

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**Homogeneous Representation**

every rigid-body transformation can be represented as a rotation followed by a translation in the same frame as a 4x4 matrix where R is a 3x3 rotation matrix and d is a 3x1 translation vector 4/14/2017

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**Homogeneous Representation**

in some frame i points vectors 4/14/2017

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**Inverse Transformation**

the inverse of a transformation undoes the original transformation if then 4/14/2017

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Day 06 Forward Kinematics 4/14/2017

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Transform Equations {2} {1} {0} {4} {3} 4/14/2017

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Transform Equations give expressions for: 4/14/2017

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Transform Equations {1} {0} {3} {2} 4/14/2017

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Transform Equations how can you find 4/14/2017

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**Links and Joints n joints, n + 1 links link 0 is fixed (the base)**

joint i connects link i – 1 to link i link i moves when joint i is actuated joint n joint 3 joint 4 joint n-1 link 3 joint 1 link 2 link n-1 link n link 1 joint 2 link 0 4/14/2017

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Forward Kinematics given the joint variables and dimensions of the links what is the position and orientation of the end effector? p0 ? a2 q2 y0 a1 q1 x0 Forward Kinematics 4/14/2017

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Forward Kinematics because the base frame and frame 1 have the same orientation, we can sum the coordinates to find the position of the end effector in the base frame (a1 cos q1 + a2 cos (q1 + q2), a1 sin q1 + a2 sin (q1 + q2) ) y1 a2 ( a2 cos (q1 + q2), a2 sin (q1 + q2) ) q2 y0 q1 a1 x1 q1 ( a1 cos q1 , a1 sin q1 ) x0 Forward Kinematics 4/14/2017

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**Forward Kinematics from Day 02 p0 = (a1 cos q1 + a2 cos (q1 + q2),**

a1 sin q1 + a2 sin (q1 + q2) ) y2 x2 x2 = (cos (q1 + q2), sin (q1 + q2) ) a2 y2 = (-sin (q1 + q2), cos (q1 + q2) ) q2 y0 q1 a1 q1 x0 Forward Kinematics 4/14/2017

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Frames y2 x2 y1 a2 q2 y0 x1 a1 q1 x0 Forward Kinematics 4/14/2017

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Forward Kinematics using transformation matrices 4/14/2017

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Day 07 Denavit-Hartenberg 4/14/2017

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**Links and Joints n joints, n + 1 links link 0 is fixed (the base)**

joint i connects link i – 1 to link i link i moves when joint i is actuated joint n joint 3 joint 4 joint n-1 link 3 joint 1 link 2 link n-1 link n link 1 joint 2 link 0 4/14/2017

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**Forward Kinematics attach a frame {i} to link i**

all points on link i are constant when expressed in {i} if joint i is actuated then frame {i} moves relative to frame {i - 1} motion is described by the rigid transformation the state of joint i is a function of its joint variable qi (i.e., is a function of qi) this makes it easy to find the last frame with respect to the base frame 4/14/2017

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**Forward Kinematics more generally**

the forward kinematics problem has been reduced to matrix multiplication 4/14/2017

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Forward Kinematics Denavit J and Hartenberg RS, “A kinematic notation for lower- pair mechanisms based on matrices.” Trans ASME J. Appl. Mech, 23:215–221, 1955 described a convention for standardizing the attachment of frames on links of a serial linkage common convention for attaching reference frames on links of a serial manipulator and computing the transformations between frames 4/14/2017

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Denavit-Hartenberg 4/14/2017

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**Denavit-Hartenberg notice the form of the rotation component**

this does not look like it can represent arbitrary rotations can the DH convention actually describe every physically possible link configuration? 4/14/2017

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Denavit-Hartenberg yes, but we must choose the orientation and position of the frames in a certain way (DH1) (DH2) claim: if DH1 and DH2 are true then there exists unique numbers 4/14/2017

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Denavit-Hartenberg 4/14/2017

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Denavit-Hartenberg proof: on blackboard in class 4/14/2017

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Day 08 Denavit-Hartenberg 4/14/2017

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**Denavit-Hartenberg Forward Kinematics**

RPP cylindrical manipulator 4/14/2017

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**Denavit-Hartenberg Forward Kinematics**

How do we place the frames? 4/14/2017

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**Step 1: Choose the z-axis for each frame**

recall the DH transformation matrix 4/14/2017

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**Step 1: Choose the z-axis for each frame**

axis of actuation for joint i+1 link i link i link i+1 link i+1 joint i+1 joint i+1 4/14/2017

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**Step 1: Choose the z-axis for each frame**

Warning: the picture is deceiving. We do not yet know the origin of the frames; all we know at this point is that each zi points along a joint axis 4/14/2017

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**Step 2: Establish frame {0}**

place the origin o0 anywhere on z0 often the choice of location is obvious choose x0 and y0 so that {0} is right-handed often the choice of directions is obvious 4/14/2017

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**Step 2: Establish frame {0}**

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

using frame {i-1} construct frame {i} DH1: xi is perpendicular to zi-1 DH2: xi intersects zi-1 3 cases to consider depending on the relationship between zi-1 and zi 4/14/2017

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

Case 1 zi-1 and zi are not coplanar (skew) ai angle from zi-1 to zi measured about xi shortest line between and (out of page) point of intersection 4/14/2017

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

Case 2 zi-1 and zi are parallel ( ai = 0 ) notice that this choice results in di = 0 point of intersection 4/14/2017

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

Case 3 zi-1 and zi intersect ( ai = 0 ) point of intersection (out of page) 4/14/2017

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

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**Step 3: Iteratively construct {1}, {2}, ... {n-1}**

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**Step 4: Place the end effector frame**

“sliding” “normal” “approach” 4/14/2017

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**Step 4: Place the end effector frame**

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**Step 5: Find the DH parameters**

ai : distance between zi-1 and zi measured along xi ai : angle from zi-1 and zi measured about xi di : distance between oi-1 to the intersection of xi and zi-1 measured along zi-1 qi : angle from xi-1 and xi measured about zi-1 4/14/2017

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**Step 5: Find the DH parameters**

Link ai di qi 1 d1 q1* 2 -90 d2* 3 d3* * joint variable 4/14/2017

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**More Denavit-Hartenberg Examples**

Day 09 More Denavit-Hartenberg Examples 4/14/2017

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**Step 5: Find the DH parameters**

Link ai di qi 1 d1 q1* 2 -90 d2* 3 d3* * joint variable 4/14/2017

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**Step 6: Compute the transformation**

once the DH parameters are known, it is easy to construct the overall transformation Link ai di qi 1 d1 q1* 2 -90 d2* 3 d3* * joint variable 4/14/2017

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**Step 6: Compute the transformation**

Link ai di qi 1 d1 q1* 2 -90 d2* 3 d3* * joint variable 4/14/2017

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**Step 6: Compute the transformation**

Link ai di qi 1 d1 q1* 2 -90 d2* 3 d3* * joint variable 4/14/2017

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**Step 6: Compute the transformation**

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Spherical Wrist 4/14/2017

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Spherical Wrist 4/14/2017

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Spherical Wrist: Step 1 4/14/2017

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Spherical Wrist: Step 2 4/14/2017

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Spherical Wrist: Step 2 4/14/2017

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Spherical Wrist: Step 4 4/14/2017

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**Step 5: DH Parameters Link ai di qi 4 -90 q4* 5 90 q5* 6 d6 q6***

-90 q4* 5 90 q5* 6 d6 q6* * joint variable 4/14/2017

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**Step 6: Compute the transformation**

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RPP + Spherical Wrist 4/14/2017

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RPP + Spherical Wrist 4/14/2017

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**Stanford Manipulator + Spherical Wrist**

Link ai di qi 1 -90 q1* 2 90 d2 q2* 3 d3* 4 q4* 5 q5* 6 d6 q6* * joint variable 4/14/2017

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**SCARA + 1DOF Wrist Link ai di qi 1 a1 d1 q1* 2 a2 180 q2* 3 d3* 4 d4**

d1 q1* 2 a2 180 q2* 3 d3* 4 d4 q4* * joint variable 4/14/2017

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