ECE5463: Introduction to Robotics Lecture Note 13: Dynamics of Open Chains: Newton Euler Approach Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018 Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 22 Outline Introduction • Newton-Euler Inverse Dynamics of Open Chains • Forward Dynamics of Open Chains • Outline Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 22 Lagrangian vs. Newton-Euler Methods • We have introduced the Lagrangian approach in the previous lecture. This lecture will introduce the Newton-Euler method • Recall the differences between the two methods: Lagrangian Formulation Newton-Euler Formulation - Energy-based method - Balance of forces/torques - Dynamic equations in closed - Dynamic equations in form numeric/recursive form - Often used for study of dynamic - Often used for numerical properties and analysis of solution of forward/inverse control methods dynamics Introduction Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 3 / 22 Main Ideas for Inverse Dynamics • Goal: Given state (θ; θ_) and acelleration θ¨, compute the required τ Rn 2 τ = M(θ)θ¨ + h(θ; θ_) • We will not drive the analytical expressions for the overall M(θ) and h(θ; θ_). • Instead, we aim to write dynamic equations separately for each link/body. T total _ Chapter 8. Dynamicsi of= Openi Chainsi [ad i ] ( i i) 293 F G V − V G V joint axis i ˙ Vi Vi i { } Fi AdT ( ) − Ti+1,i Fi+1 Figure 8.6: Free-body diagram illustrating the moments and forces exerted on link i. Introduction Substituting this resultLecture into 13 Equation (ECE5463 Sp18)(8.46), we get Wei Zhang(OSU) 4 / 22 ˙ ¨ ˙ ˙ i = iθi + [AdTi,i 1 ] i 1 + [ad i ] iθi, (8.47) V A − V − V A i.e., the acceleration of link i is the sum of three components: a component due to the joint acceleration θ¨i, a component due to the acceleration of link i 1 expressed in i , and a velocity-product component. − Once we have{ } determined all the link twists and accelerations moving out- ward from the base, we can calculate the joint torques or forces by moving inward from the tip. The rigid-body dynamics (8.40) tells us the total wrench that acts on link i given and ˙ . Furthermore, the total wrench acting on Vi Vi link i is the sum of the wrench i transmitted through joint i and the wrench applied to the link through jointFi + 1 (or, for link n, the wrench applied to the link by the environment at the end-effector frame n + 1 ), expressed in the frame i. Therefore, we have the equality { } ˙ T T i i ad ( i i) = i AdT ( i+1); (8.48) G V − Vi G V F − i+1,i F see Figure 8.6. Solving from the tip toward the base, at each link i we solve for the only unknown in Equation (8.48): . Since joint i has only one degree Fi of freedom, five dimensions of the six-vector i are provided “for free” by the structure of the joint, and the actuator onlyF has to provide the scalar force or torque in the direction of the joint’s screw axis: τ = T . (8.49) i Fi Ai Equation (8.49) provides the torques required at each joint, solving the inverse dynamics problem. May 2017 preprint of Modern Robotics, Lynch and Park, Cambridge U. Press, 2017. http://modernrobotics.org Main Ideas for Inverse Dynamics (Continued) • Note that i can be easily computed for each link. Thus, the main task is to find ; _ Gand the total wrench for each link in terms of the given (θ; θ;_ θ¨). Vi Vi • This can be done recursively: - The velocity Vi of link i is determined by Vi−1 of link i − 1 and the joint rate θ_i of joint i - The acceleration V_ i of link i is determined by V_ i−1 of link i − 1 and the acceleration θ¨i of joint i - The total wrench of link i can be determined by gravity and the wrench of adjacent links using the principle of action and reaction: Wrench applied by body i to body i+1 = − Wrench applied by body i+1 to body i Introduction Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 5 / 22 Outline Introduction • Newton-Euler Inverse Dynamics of Open Chains • Forward Dynamics of Open Chains • Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 6 / 22 Some Notations (1/3) • For each link i = 1; : : : ; n, Frame i is attached to its center of mass. f g • The base frame is denoted 0 f g • The end-effector frame is denoted n + 1 . Its configuration in n is fixed. f g f g • M : Configuration of j in i when the robot is at home position i;j f g f g • T (θ): Configuration of j in i when the robot is at θ position i;j f g f g Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 7 / 22 Some Notations (2/3) • All the following quantities are expressed in frame i f g - Vi: Twist of link fig - ¯mi: mass; Ii: rotational inertia matrix; Ii 0 - Gi = : Spatial inertia matrix 0 ¯miI Chapter 8. Dynamics of Open Chains 293 - Ai: Screw axis of joint i, expressed in fig joint axis i ˙ Vi i - Fi = (mi; fi): wrench transmitted through V joint i to link fig i { } Fi AdT ( ) − Ti+1,i Fi+1 Figure 8.6: Free-body diagram illustrating the moments and forces exerted on link i. Newton-Euler Inverse Dynamics Substituting this resultLecture into 13 Equation (ECE5463 Sp18)(8.46), we get Wei Zhang(OSU) 8 / 22 ˙ ¨ ˙ ˙ i = iθi + [AdTi,i 1 ] i 1 + [ad i ] iθi, (8.47) V A − V − V A i.e., the acceleration of link i is the sum of three components: a component due to the joint acceleration θ¨i, a component due to the acceleration of link i 1 expressed in i , and a velocity-product component. − Once we have{ } determined all the link twists and accelerations moving out- ward from the base, we can calculate the joint torques or forces by moving inward from the tip. The rigid-body dynamics (8.40) tells us the total wrench that acts on link i given and ˙ . Furthermore, the total wrench acting on Vi Vi link i is the sum of the wrench i transmitted through joint i and the wrench applied to the link through jointFi + 1 (or, for link n, the wrench applied to the link by the environment at the end-effector frame n + 1 ), expressed in the frame i. Therefore, we have the equality { } ˙ T T i i ad ( i i) = i AdT ( i+1); (8.48) G V − Vi G V F − i+1,i F see Figure 8.6. Solving from the tip toward the base, at each link i we solve for the only unknown in Equation (8.48): . Since joint i has only one degree Fi of freedom, five dimensions of the six-vector i are provided “for free” by the structure of the joint, and the actuator onlyF has to provide the scalar force or torque in the direction of the joint’s screw axis: τ = T . (8.49) i Fi Ai Equation (8.49) provides the torques required at each joint, solving the inverse dynamics problem. May 2017 preprint of Modern Robotics, Lynch and Park, Cambridge U. Press, 2017. http://modernrobotics.org Some Notations (3/3) • Some simple facts: −1 −1 - Mi−1;i = Mi−1Mi and Mi;i−1 = Mi Mi−1 h i - Ai = AdM 1 Si, where Si is the screw axis of joint i (relative to f0g frame) i− when θ = 0 [A ]θ −[A ]θ - Ti−1;i(θ) = Ti−1;i(θi) = Mi−1;ie i i and Ti;i−1(θi) = e i i Mi;i−1 Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 9 / 22 Forward Propagation of Twist (1/3) • Goal: Given i 1 and (θi; θ_i), we want to find i V − V • Generally speaking, there are two key elements in defining a rigid body velocity: - The inertia frame relative to which the motion is considered - The representation frame in which the coordinate of the velocity vector is determined. • So far, we have mainly played with the representation frames, while the motion is always considered to be relative to the fixed s frame. f g • Sometimes, it is convenient consider a relative velocity with respect to a moving frame. Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 10 / 22 Forward Propagation of Twist (2/3) b=c • Define a as the velocity (twist) of b relative to c , expressed in a . V f g f g f g • Connecting to our previous shorthand notations: = b=s; = b=s; = i=s Vs Vs Vb Vb Vi Vi • With Newtonian approximation: we have c=s = b=s + c=b V∗ V∗ V∗ - The "*" can be replaced with fsg or fbg or any other stationary frames. - Note the representation frame must be the same for all the three twists. Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 11 / 22 Forward Propagation of Twist (3/3) • Now consider velocity of link i: we will have i=s i=i 1 i 1=s = − + − Vi Vi Vi i=i 1 • Note that i − can be viewed as the body twist of a 1-joint robot where V i=i 1 i-1 is viewed as the “fixed” frame − = θ_ f g )V i Ai i • Moreover, with our shorthand notation: i−1=s i−1=s Vi = AdTi;i 1 Vi−1 = AdTi;i 1 Vi−1 − − • Therefore, velocity (twist) can be propagated by: _ i = iθi + AdTi;i 1 i 1 (1) V A − V − Newton-Euler Inverse Dynamics Lecture 13 (ECE5463 Sp18) Wei Zhang(OSU) 12 / 22 Forward Propagation of Acceleration (1/2) • Goal: Given _ i 1, i 1 and (θi; θ_i; θ¨i), we want to find _ i, V − V − V • We first mention a result about the derivative of the adjoint operator.