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ECE5463: Introduction to Lecture Note 11: of a Single

Prof. Wei Zhang

Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA

Spring 2018

Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 1 / 24 Outline

• Kinetic of a Rigid Body

• Rotational Inertia Matrix

• Euler Equation

• Twist-Wrench Formulation of Rigid-Body Dynamics

Outline Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 2 / 24 Robot Dynamic Model Can be Complicated

• Dynamic model of PUMA 560 Arm:

Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 3 / 24

• Consider a point ¯m with {s}-frame coordinate p(t), its kinetic energy is given by 1 K = ¯mp˙2 2

• Note: m denotes (vector) and ¯m denotes mass (scalar).

• Question: given a moving rigid body with configuration T (t)=(R(t),p(t)), how to compute its kinetic energy? - Rigid body consists of infinitely many “particles” with different {s}-frame  1 2 K≈ ¯mip˙i 2 i

- Velocities of particles p˙i are caused by the rigid body (twist)

- The overall kinetic energy should depend on the rigid body velocity as well as the geometry and mass distribution of the body

Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 4 / 24 Recall: Rigid Body Velocity Given rigid body T (t)=(R(t),p(t)): • Spatial twist:

• Body twist:

Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 5 / 24 Recall: Rigid Body Velocity

• Consider a particle i on the body with {b}-frame coordinate ri and {s}-frame coordinate pi - Velocity of particle i:

- of particle i:

- Velocity of the origin of {b}:

Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 6 / 24 Rigid Body Kinetic Energy

• Kinetic Energy: GivenarigidbodyT (t)=(R(t),p(t)) with body twist Vb =(ωb,vb). Suppose the {b}-frame origin coincides with the center of mass of the body. Then its kinetic energy is given by: 1 2 1 T T K = ¯mvb + ωb Ibωb, with Ib = ρ(r)[r] [r]dV 2 2 B

where Ib is the rotational inertia matrix in body frame

Derivation:D Divide the body into small point , where point i has mass ¯mi, {b}-frame coordinate ri,and{s}-frame coordinate pi

Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 7 / 24 Derivation of Kinetic Energy (Continued)

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Kinetic Energy Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 8 / 24 Outline

• Kinetic Energy of a Rigid Body

• Rotational Inertia Matrix

• Newton Euler Equation

• Twist-Wrench Formulation of Rigid-Body Dynamics

Rotational Inertia Matrix Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 9 / 24 Rotational Inertia Matrix in Body Frame T Ib  ρ(r)[r] [r]dV B

• Individual entries of Ib: ⎡ ⎤ Ixx Ixy Ixz ⎣ ⎦ Ib = Iyx Iyy Iyz Izx Izy Izz

where 2 2 2 2 Ixx = (y + z )ρ(x, y, z)dV, Iyy = (x + z )ρ(x, y, z)dV B B 2 2 Izz = (x + y )ρ(x, y, z)dV, Ixy = Iyx = − xyρ(x, y, z)dV B B Ixz = Izx = − xzρ(x, y, z)dV Iyz = Izy = − yzρ(x, y, z)dV B B

• If the body has a uniform density, then Ib is determined exclusively by the shape of the rigid body

Rotational Inertia Matrix Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 10 / 24 Principal Axes of Inertia

Let v1,v2,v3 and λ1,λ2,λ3 be the eigenvectors and eigenvalues of Ib, respectively. They are called the principal axes of inertia

• The principal axes of inertia are in the directions of v1,v2,v3

• The principal moments of inertia about these axes are λ1,λ2,λ3

• All the eigenvalues are nonnegative. The largest one maximizes the among all the axes passing through the center of mass of the body.

• If the principal axes of inertia are aligned with the axes of {b},the off-diagonal terms of Ib are all zero.

Rotational Inertia Matrix Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 11 / 24 Examples of Inertia Matrix

ˆz ˆz ˆz

h c h r yˆ yˆ xˆ a b yˆ xˆ xˆ w

rectangular parallelepiped: circular cylinder: ellipsoid: volume = abc,volume=πr2h, volume = 4πabc/3, 2 2 2 2 2 2 Ixx = m(w + h )/12, Ixx = m(3r + h )/12, Ixx = m(b + c )/5, 2 2 2 2 2 2 Iyy = m( + h )/12, Iyy = m(3r + h )/12, Iyy = m(a + c )/5, 2 2 2 2 2 Izz = m( + w )/12 Izz = mr /2 Izz = m(a + b )/5

The principal axes and the inertia about the principal axes for uniform-density bodies

Rotational Inertia Matrix Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 12 / 24 Inertia Matrix in a Different Frame

• Consider another frame {c} with relative orientation Rbc

• The origin of both frames is located at the CoM of the body. The rotational { } I ρ r r T r dV inertia matrix in c frame is defined as c = B ( c)[ c] [ c]

T • Kinetic energy is independent of reference frames ⇒ Ic = RbcIbRbc

• Steiner’s Theorem: The inertia matrix Iq about a frame aligned with {b}, but at a point q =(qx,qy,qz) in {b}, is related to Ib by T T Iq = Ib + ¯m q qII − qq

Rotational Inertia Matrix Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 13 / 24 Outline

• Kinetic Energy of a Rigid Body

• Rotational Inertia Matrix

• Newton Euler Equation

• Twist-Wrench Formulation of Rigid-Body Dynamics

Newton Euler Equation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 14 / 24 Newton Euler Equation

• Recall that for a point mass ¯m with a fixed-frame coordinate p(t),Newton’s second law of : f = ¯mp¨(t) • A rigid body consists of infinitely many point masses. The collective motion of these particles depend on the linear and rotational velocities of the body, and on the total and moment acting on the body. • Euler-Newton Equation of Motion: Given rigid body T (t)=(R(t),p(t)) with rotational inertia matrix Ib and body twist Vb =(ωb,vb): mb = Ibω˙b + ωb ×Ibωb (1) fb = ¯mv˙b + ωb × ¯mvb

- ¯m: mass of the body; assume origin of {b} =CoM

- fb,mb: total force and moment (expressed in {b})actingonthebody

- ¯mvb:isthelinear of the body

- Ibωb:istheangular momentum of the body

Newton Euler Equation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 15 / 24 Derivation of Newton Euler Equation

Newton Euler Equation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 16 / 24 Derivation of Newton Euler Equation (Continued...)

Newton Euler Equation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 17 / 24

Outline

• Kinetic Energy of a Rigid Body

• Rotational Inertia Matrix

• Newton Euler Equation

• Twist-Wrench Formulation of Rigid-Body Dynamics

Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 18 / 24 Lie Bracket

• Lie Bracket: Given two twists V1 =(ω1,v1) and V2 =(ω2,v2), the Lie V V V bracket of 1 and 2, written as [adV1 ] 2, is defined as follows: [ω1]0 ω2 6 [adV1 ] V2 = ∈ R [v1][ω1] v2 [ω]0 where [adV ]  for any V =(ω, v) ∈ se(3) [v][ω]

• Lie Bracket can be viewed as a generalization of the cross-product operation oftwovectorstotwotwists

• Given a twist V =(ω, v) and a wrench F =(m, f), we define the mapping:

T T [ω]0 m [adV ] F = [v][ω] f

Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 19 / 24 Twist-Wrench Formulation

• Rigid body with body twist Vb =(ωb,vb) and body wrench Fb =(mb,fb)

6×6 Ib 0 • Spatial inertia matrix Gb ∈ R : Gb = 0 ¯mI

6 • Spatial momentum Pb ∈ R : Pb = GbVb

• 1 T Kinetic energy: K = 2 Vb GbVb

Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 20 / 24 Twist-Wrench Formulation

• Newton-Euler Equation (1) can be written in twist-wrench form:

F G V˙ − T P G V˙ − T G V b = b b [adVb ] b = b b [adVb ] b b

Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 21 / 24 Dynamics in Other Frames

• Our derivation of dynamics relies on using CoM {b} frame. We can also write dynamics in another frame, say {a}, with relative configuration Tba

• 1 T 1 T Kinetic energy is independent of reference frame: 2 Vb GbVb = 2 Va GaVa

• This implies that the spatial inertia matrix Ga is related to Gb by

G T G a =[AdTba ] b [AdTba ]

• One can show the Newton-Euler equation (1) can be written equivalently in frame {a} as: F G V˙ − T G V a = a a [adVa ] a a

Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 22 / 24 More Discussions

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Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 23 / 24 More Discussions

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Twist-Wrench Formulation Lecture 11 (ECE5463 Sp18) Wei Zhang(OSU) 24 / 24