[足式机器人]Part4 南科大高等机器人控制课 Ch02 Rigid Body Configuration and Velocity

LiongLoure2023-12-10 18:16

本文仅供学习使用
本文参考：
B站：CLEAR_LAB
笔者带更新-运动学
课程主讲教师：
Prof. Wei Zhang

南科大高等机器人控制课 Ch02 Rigid Body Configuration and Velocity

  • [1. Rigid Body Configuration](#1. Rigid Body Configuration)
    • [1.1 Special Orthogonal Group](#1.1 Special Orthogonal Group)
    • [1.2 Use of Rotation Matrix](#1.2 Use of Rotation Matrix)
    • [1.3 Homogeneous Transformation Matrix](#1.3 Homogeneous Transformation Matrix)
  • [2. Rigid Body Velocity（Twist）](#2. Rigid Body Velocity（Twist）)
    • [2.1 Rigid Body Velocity: Spatial Velocity (Twist)](#2.1 Rigid Body Velocity: Spatial Velocity (Twist))
    • [2.2 Spatial Velocity Representation in a Reference Frame](#2.2 Spatial Velocity Representation in a Reference Frame)
    • [2.3 Change Reference Frame for Twist](#2.3 Change Reference Frame for Twist)
  • [3. Geometric Aspect of Twist: Screw Motion](#3. Geometric Aspect of Twist: Screw Motion)
    • [3.1 Screw Motion : Definition](#3.1 Screw Motion : Definition)
    • [3.2 From Screw Motion to Twist](#3.2 From Screw Motion to Twist)
    • [3.3 From Twist to Screw Motion](#3.3 From Twist to Screw Motion)
    • [3.4 Screw Reoersentation of a Twist](#3.4 Screw Reoersentation of a Twist)
  • [4. Extra Note : Tutorial on Twist/spatial Velocity and Screw Axis](#4. Extra Note : Tutorial on Twist/spatial Velocity and Screw Axis)
    • [4.1 What is Spatial Velocity and Twist](#4.1 What is Spatial Velocity and Twist)
    • [4.2 What is Screw Motion and Axis?](#4.2 What is Screw Motion and Axis?)

1. Rigid Body Configuration

  • Free Vector

Free Vector: geometric quantity with length and direction

Given a reference frame, v ⃗ \vec{v} v can be move to a position such that the base of the arrow is at the origin without changing the orientation. Then the vector v ⃗ \vec{v} v can be represented by its coordinates v ⃗ \vec{v} v in the reference frame

v ⃗ \vec{v} v donated the physical quantity while v ⃗ A \vec{v}^A v A donate its coordinate wrt frame { A } \left\{ A \right\} {A}

Frame: coordinate system based on basis vectors------ { A : i ^ A , j ^ A , k ^ A } \left\{ A:\hat{i}^A,\hat{j}^A,\hat{k}^A \right\} {A:i^A,j^A,k^A}------3 coordinate vectors (unit length) i ^ A , j ^ A , k ^ A \hat{i}^A,\hat{j}^A,\hat{k}^A i^A,j^A,k^A and an origin

i ^ A , j ^ A , k ^ A \hat{i}^A,\hat{j}^A,\hat{k}^A i^A,j^A,k^A mutually orthogonal
i ^ A × j ^ A = k ^ A \hat{i}^A\times \hat{j}^A=\hat{k}^A i^A×j^A=k^A------right hand rule

  • Point

Point : p p p denotes a point in the physical space

{ A } \left\{ A \right\} {A} point p p p can be represented by a vector from frame origin to p p p
R ⃗ P A \vec{R}_{\mathrm{P}}^{A} R PA denotes the coordinate of a point p p p wrt frame { A } \left\{ A \right\} {A}

此处使用了笔者习惯的表达方式，所以并不会出现不同坐标系下表达的向量不可相加的情况（本质并非坐标参数的相加），这种表达方式很多，本质都是为了简化直观向量表达的同时不产生歧义

  • Cross Product

Cross Product or vector product of a ⃗ ∈ R 3 , b ⃗ ∈ R 3 \vec{a}\in \mathbb{R} ^3,\vec{b}\in \mathbb{R} ^3 a ∈R3,b ∈R3 is defined as
a ⃗ × b ⃗ = [ I ^ J ^ K ^ ] T [ a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ] = [ I ^ J ^ K ^ ] T [ 0 − a 3 a 2 a 3 0 − a 1 − a 2 a 1 0 ] [ b 1 b 2 b 3 ] = [ I ^ J ^ K ^ ] T a ⃗ ~ [ b 1 b 2 b 3 ] \vec{a}\times \vec{b}=\left[ \begin{array}{c} \hat{I}\\ \hat{J}\\ \hat{K}\\ \end{array} \right] ^{\mathrm{T}}\left[ \begin{array}{c} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1\\ \end{array} \right] =\left[ \begin{array}{c} \hat{I}\\ \hat{J}\\ \hat{K}\\ \end{array} \right] ^{\mathrm{T}}\left[ \begin{matrix} 0& -a_3& a_2\\ a_3& 0& -a_1\\ -a_2& a_1& 0\\ \end{matrix} \right] \left[ \begin{array}{c} b_1\\ b_2\\ b_3\\ \end{array} \right] =\left[ \begin{array}{c} \hat{I}\\ \hat{J}\\ \hat{K}\\ \end{array} \right] ^{\mathrm{T}}\tilde{\vec{a}}\left[ \begin{array}{c} b_1\\ b_2\\ b_3\\ \end{array} \right] a ×b = I^J^K^ T a2b3−a3b2a3b1−a1b3a1b2−a2b1 = I^J^K^ T 0a3−a2−a30a1a2−a10 b1b2b3 = I^J^K^ Ta ~ b1b2b3

a ⃗ ~ = − a ⃗ ~ T \tilde{\vec{a}}=-\tilde{\vec{a}}^{\mathrm{T}} a ~=−a ~T (called skew stmmetric )
a ⃗ ~ b ⃗ ~ − b ⃗ ~ a ⃗ ~ = a ⃗ × b ⃗ ~ \tilde{\vec{a}}\tilde{\vec{b}}-\tilde{\vec{b}}\tilde{\vec{a}}=\widetilde{\vec{a}\times \vec{b}} a ~b ~−b ~a ~=a ×b Jocabi's Idenetity

  • Rotation Matrix

Rotation Matrix: specifies orientation of one frame relative to another

A valid rotation matrx [ Q B A ] \left[ Q_{\mathrm{B}}^{A} \right] [QBA] satisfies : [ Q B A ] [ Q B A ] T = E , det ⁡ ( [ Q B A ] ) = 1 \left[ Q_{\mathrm{B}}^{A} \right] \left[ Q_{\mathrm{B}}^{A} \right] ^{\mathrm{T}}=E,\det \left( \left[ Q_{\mathrm{B}}^{A} \right] \right) =1 [QBA][QBA]T=E,det([QBA])=1

1.1 Special Orthogonal Group

Special Orthogonal Group : Space of Rotation Matrices in R n \mathbb{R} ^n Rn is defined as
S O ( n ) = { [ Q B A ] ∈ R n × n : [ Q B A ] [ Q B A ] T = E , det ⁡ ( [ Q B A ] ) = 1 } SO\left( n \right) =\left\{ \left[ Q_{\mathrm{B}}^{A} \right] \in \mathbb{R} ^{n\times n}:\left[ Q_{\mathrm{B}}^{A} \right] \left[ Q_{\mathrm{B}}^{A} \right] ^{\mathrm{T}}=E,\det \left( \left[ Q_{\mathrm{B}}^{A} \right] \right) =1 \right\} SO(n)={[QBA]∈Rn×n:[QBA][QBA]T=E,det([QBA])=1}

S O ( n ) SO\left( n \right) SO(n) is a group. We are primarily interested in S O ( 3 ) SO\left( 3 \right) SO(3) and S O ( 2 ) SO\left( 2 \right) SO(2), rotation groups of R 3 \mathbb{R} ^3 R3 and R 2 \mathbb{R} ^2 R2 , respectively.

Group is a set G G G, together with an operation ∙ \bullet ∙, satisfying the following group axioms/'æksɪəm/公理:

  • Closure: a ∈ G , b ∈ G ⇒ a ∙ b ∈ G a\in G,b\in G\Rightarrow a\bullet b\in G a∈G,b∈G⇒a∙b∈G
  • Assocaitivity: ( a ∙ b ) ∙ c = a ∙ ( b ∙ c ) , ∀ a , b , c ∈ G \left( a\bullet b \right) \bullet c=a\bullet \left( b\bullet c \right) ,\forall a,b,c\in G (a∙b)∙c=a∙(b∙c),∀a,b,c∈G
  • Identity element: ∃ e ∈ G \exists e\in G ∃e∈G such that e ∙ a = a e\bullet a=a e∙a=a , for all a ∈ G a\in G a∈G
  • Inverse element: For each a ∈ G a\in G a∈G, there is a b ∈ G b\in G b∈G such that a ∙ b = b ∙ a = e a\bullet b=b\bullet a=e a∙b=b∙a=e, where e e e is the identity element

1.2 Use of Rotation Matrix

  • Representing an orientation ------ from definition
    将原矢量进行旋转变换，得到该坐标系下新矢量的坐标投影参数：
    R ⃗ p ′ F = [ Q B A ] R ⃗ p F \vec{R}{\mathrm{p}^{\prime}}^{F}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{p}}^{F} R p′F=[QBA]R pF
  • Changing the reference frame
    对坐标系进行转换，基于坐标系 { B } \left\{ B \right\} {B}中的该矢量的坐标投影参数 R ⃗ p B \vec{R}{\mathrm{p}}^{B} R pB，得到该矢量在坐标系 { A } \left\{ A \right\} {A}中的坐标投影参数 R ⃗ p A \vec{R}{\mathrm{p}}^{A} R pA：
    R ⃗ p A = [ Q B A ] R ⃗ p B \vec{R}{\mathrm{p}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{p}}^{B} R pA=[QBA]R pB

i ⃗ B j ⃗ B k ⃗ B \] T \[ P 1 B P 2 B P 3 B \] = \[ i ⃗ A j ⃗ A k ⃗ A \] T \[ P 1 A P 2 A P 3 A \] ⇒ ( \[ Q B A \] T \[ i ⃗ A j ⃗ A k ⃗ A \] ) T \[ P 1 B P 2 B P 3 B \] = \[ i ⃗ A j ⃗ A k ⃗ A \] T \[ P 1 A P 2 A P 3 A \] ⇒ \[ i ⃗ A j ⃗ A k ⃗ A \] T \[ Q B A \] \[ P 1 B P 2 B P 3 B \] = \[ i ⃗ A j ⃗ A k ⃗ A \] T \[ P 1 A P 2 A P 3 A \] ⇒ \[ Q B A \] \[ P 1 B P 2 B P 3 B \] = \[ P 1 A P 2 A P 3 A \] = \[ P ′ 1 B P ′ 2 B P ′ 3 B \] \\left\[ \\begin{array}{c} \\vec{i}\^B\\\\ \\vec{j}\^B\\\\ \\vec{k}\^B\\\\ \\end{array} \\right\] \^{\\mathrm{T}}\\left\[ \\begin{array}{c} P_{1}\^{\\mathrm{B}}\\\\ P_{2}\^{\\mathrm{B}}\\\\ P_{3}\^{\\mathrm{B}}\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} \\vec{i}\^A\\\\ \\vec{j}\^A\\\\ \\vec{k}\^A\\\\ \\end{array} \\right\] \^{\\mathrm{T}}\\left\[ \\begin{array}{c} P_{1}\^{A}\\\\ P_{2}\^{A}\\\\ P_{3}\^{A}\\\\ \\end{array} \\right\] \\\\ \\Rightarrow \\left( \\left\[ Q_{\\mathrm{B}}\^{A} \\right\] \^{\\mathrm{T}}\\left\[ \\begin{array}{c} \\vec{i}\^A\\\\ \\vec{j}\^A\\\\ \\vec{k}\^A\\\\ \\end{array} \\right\] \\right) \^{\\mathrm{T}}\\left\[ \\begin{array}{c} P_{1}\^{\\mathrm{B}}\\\\ P_{2}\^{\\mathrm{B}}\\\\ P_{3}\^{\\mathrm{B}}\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} \\vec{i}\^A\\\\ \\vec{j}\^A\\\\ \\vec{k}\^A\\\\ \\end{array} \\right\] \^{\\mathrm{T}}\\left\[ \\begin{array}{c} P_{1}\^{A}\\\\ P_{2}\^{A}\\\\ P_{3}\^{A}\\\\ \\end{array} \\right\] \\\\ \\Rightarrow \\left\[ \\begin{array}{c} \\vec{i}\^A\\\\ \\vec{j}\^A\\\\ \\vec{k}\^A\\\\ \\end{array} \\right\] \^{\\mathrm{T}}\\left\[ Q_{\\mathrm{B}}\^{A} \\right\] \\left\[ \\begin{array}{c} P_{1}\^{\\mathrm{B}}\\\\ P_{2}\^{\\mathrm{B}}\\\\ P_{3}\^{\\mathrm{B}}\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} \\vec{i}\^A\\\\ \\vec{j}\^A\\\\ \\vec{k}\^A\\\\ \\end{array} \\right\] \^{\\mathrm{T}}\\left\[ \\begin{array}{c} P_{1}\^{A}\\\\ P_{2}\^{A}\\\\ P_{3}\^{A}\\\\ \\end{array} \\right\] \\\\ \\Rightarrow \\left\[ Q_{\\mathrm{B}}\^{A} \\right\] \\left\[ \\begin{array}{c} P_{1}\^{\\mathrm{B}}\\\\ P_{2}\^{\\mathrm{B}}\\\\ P_{3}\^{\\mathrm{B}}\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} P_{1}\^{A}\\\\ P_{2}\^{A}\\\\ P_{3}\^{A}\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} {P\^{\\prime}}_{1}\^{\\mathrm{B}}\\\\ {P\^{\\prime}}_{2}\^{\\mathrm{B}}\\\\ {P\^{\\prime}}_{3}\^{\\mathrm{B}}\\\\ \\end{array} \\right\] i Bj Bk B T P1BP2BP3B = i Aj Ak A T P1AP2AP3A ⇒ \[QBA\]T i Aj Ak A T P1BP2BP3B = i Aj Ak A T P1AP2AP3A ⇒ i Aj Ak A T\[QBA\] P1BP2BP3B = i Aj Ak A T P1AP2AP3A ⇒\[QBA\] P1BP2BP3B = P1AP2AP3A = P′1BP′2BP′3B

  • Rotating a vector or a frame (轴角变换)

Given two coordinate frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B}, the configuration of B B B relative to A A A is determined by [ Q B A ] \left[ Q_{\mathrm{B}}^{A} \right] [QBA] and R ⃗ B A \vec{R}_{\mathrm{B}}^{A} R BA

For a (free) vector R ⃗ f r e e \vec{R}{\mathrm{free}} R free, its coordinates R ⃗ f r e e A \vec{R}{free}^{A} R freeA and R ⃗ f r e e B \vec{R}{free}^{B} R freeB are related by : R ⃗ f r e e A = [ Q B A ] R ⃗ f r e e B \vec{R}{free}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{free}}^{B} R freeA=[QBA]R freeB

For a point P P P, its coordinates R ⃗ P A \vec{R}{\mathrm{P}}^{A} R PA and R ⃗ P B \vec{R}{\mathrm{P}}^{B} R PB are related by: R ⃗ P A = [ Q B A ] R ⃗ P B + R ⃗ B A \vec{R}{\mathrm{P}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}{\mathrm{P}}^{B}+\vec{R}{\mathrm{B}}^{A} R PA=[QBA]R PB+R BA(一个无聊的小陷阱)

1.3 Homogeneous Transformation Matrix

Linear relation: R ⃗ f r e e A = [ Q B A ] R ⃗ f r e e B \vec{R}{\mathrm{free}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}_{\mathrm{free}}^{B} R freeA=[QBA]R freeB------configuration of { B } \left\{ B \right\} {B} relative to { A } \left\{ A \right\} {A}

Affine relation: R ⃗ P A = [ Q B A ] R ⃗ P B + R ⃗ B A \vec{R}{\mathrm{P}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}{\mathrm{P}}^{B}+\vec{R}{\mathrm{B}}^{A} R PA=[QBA]R PB+R BA

Homogeneous Transformation Matrix: [ T B A ] \left[ T_{\mathrm{B}}^{A} \right] [TBA]
R ⃗ P A = [ Q B A ] R ⃗ P B + R ⃗ B A ⇒ [ R ⃗ P A 1 ] = [ [ Q B A ] R ⃗ B A 0 1 × 3 1 ] 4 × 4 [ R ⃗ P B 1 ] \vec{R}{\mathrm{P}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{R}{\mathrm{P}}^{B}+\vec{R}{\mathrm{B}}^{A}\Rightarrow \left[ \begin{array}{c} \vec{R}{\mathrm{P}}^{A}\\ 1\\ \end{array} \right] =\left[ \begin{matrix} \left[ Q{\mathrm{B}}^{A} \right]& \vec{R}{\mathrm{B}}^{A}\\ 0{1\times 3}& 1\\ \end{matrix} \right] {4\times 4}\left[ \begin{array}{c} \vec{R}{\mathrm{P}}^{B}\\ 1\\ \end{array} \right] R PA=[QBA]R PB+R BA⇒[R PA1]=[[QBA]01×3R BA1]4×4[R PB1]
⇒ [ T B A ] = [ [ Q B A ] R ⃗ B A 0 1 ] \Rightarrow \left[ T_{\mathrm{B}}^{A} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{B}}^{A} \right]& \vec{R}_{\mathrm{B}}^{A}\\ 0& 1\\ \end{matrix} \right] ⇒[TBA]=[[QBA]0R BA1]

Homogeneous coordinates: Given a point P ∈ R 3 P\in \mathbb{R} ^3 P∈R3, its homogeneous coordinates is given by [ R ⃗ P A ] = [ R ⃗ P A 1 ] ∈ R 4 \left[ \vec{R}{\mathrm{P}}^{A} \right] =\left[ \begin{array}{c} \vec{R}{\mathrm{P}}^{A}\\ 1\\ \end{array} \right] \in \mathbb{R} ^4 [R PA]=[R PA1]∈R4

最终简化为：

R ⃗ P A \] = \[ T B A \] \[ R ⃗ P B \] \\left\[ \\vec{R}_{\\mathrm{P}}\^{A} \\right\] =\\left\[ T_{\\mathrm{B}}\^{A} \\right\] \\left\[ \\vec{R}_{\\mathrm{P}}\^{B} \\right\] \[R PA\]=\[TBA\]\[R PB

对于向量 R ⃗ P 1 P 2 A \vec{R}_{\mathrm{P}_1\mathrm{P}_2}^{A} R P1P2A 而言，则有：

R ⃗ P 1 P 2 A \] = \[ R ⃗ P 2 A − R ⃗ P 1 A \] = \[ R ⃗ P 2 A 1 \] − \[ R ⃗ P 1 A 1 \] = \[ R ⃗ P 2 A − R ⃗ P 1 A 0 \] = \[ R ⃗ P 1 P 2 A 0 \] \\left\[ \\vec{R}_{\\mathrm{P}_1\\mathrm{P}_2}\^{A} \\right\] =\\left\[ \\vec{R}_{\\mathrm{P}_2}\^{A}-\\vec{R}_{\\mathrm{P}_1}\^{A} \\right\] =\\left\[ \\begin{array}{c} \\vec{R}_{\\mathrm{P}_2}\^{A}\\\\ 1\\\\ \\end{array} \\right\] -\\left\[ \\begin{array}{c} \\vec{R}_{\\mathrm{P}_1}\^{A}\\\\ 1\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} \\vec{R}_{\\mathrm{P}_2}\^{A}-\\vec{R}_{\\mathrm{P}_1}\^{A}\\\\ 0\\\\ \\end{array} \\right\] =\\left\[ \\begin{array}{c} \\vec{R}_{\\mathrm{P}_1\\mathrm{P}_2}\^{A}\\\\ 0\\\\ \\end{array} \\right\] \[R P1P2A\]=\[R P2A−R P1A\]=\[R P2A1\]−\[R P1A1\]=\[R P2A−R P1A0\]=\[R P1P2A0

  • Example of Homogeneous Transformation Matrix

2. Rigid Body Velocity（Twist）

Consider a rigid body with angular velocity: ω ⃗ \vec{\omega} ω (this is a free vector)

Suppose the actual rotation axis passes through a point: P P P ; Let v ⃗ P \vec{v}_{\mathrm{P}} v P be the velocity of the point P P P

  • Question: A rigid body cibraubs infinitely many points with different velocities. How to parameterize/pə'ræmɪtə,raɪz/参数化 all of their velocities?
    1.Consider an aritrary body-fixed point Q Q Q (means that the point is rigidly attached to the body, and moves with the body), we have: v ⃗ Q = v ⃗ P + ω ⃗ × R ⃗ P Q \vec{v}{\mathrm{Q}}=\vec{v}{\mathrm{P}}+\vec{\omega}\times \vec{R}{\mathrm{PQ}} v Q=v P+ω ×R PQ
    2.The velocity of an arbitrary body-fixed point depends only on ( ω ⃗ , v ⃗ P , R ⃗ P \vec{\omega},\vec{v}    {\mathrm{P}},\vec{R}_{\mathrm{P}} ω ,v P,R P) and the location of the point Q Q Q
  • Fact: The representation form is independent of the reference point P P P
  • Consider an arbitrary point S S S in space
    1. S S S may not be on the rotation axis
    2. S S S may be a stationary point in space(does not move)
      3.Let v ⃗ S \vec{v}{\mathrm{S}} v S be the velocity of the body-fixed point(rigidly attached to the body ) currently coincides with S S S(may not be body frame)
      4.We still have: v ⃗ Q = v ⃗ P + ω ⃗ × R ⃗ P Q , v ⃗ S = v ⃗ P + ω ⃗ × R ⃗ P S ⇒ v ⃗ Q = v ⃗ S − ω ⃗ × R ⃗ P S + ω ⃗ × R ⃗ P Q = v ⃗ S + ω ⃗ × R ⃗ S Q \vec{v}      {\mathrm{Q}}=\vec{v}{\mathrm{P}}+\vec{\omega}\times \vec{R}{\mathrm{PQ}},\vec{v}{\mathrm{S}}=\vec{v}{\mathrm{P}}+\vec{\omega}\times \vec{R}{\mathrm{PS}}\Rightarrow \vec{v}{\mathrm{Q}}=\vec{v}{\mathrm{S}}-\vec{\omega}\times \vec{R}{\mathrm{PS}}+\vec{\omega}\times \vec{R}{\mathrm{PQ}}=\vec{v}{\mathrm{S}}+\vec{\omega}\times \vec{R}_{\mathrm{SQ}} v Q=v P+ω ×R PQ,v S=v P+ω ×R PS⇒v Q=v S−ω ×R PS+ω ×R PQ=v S+ω ×R SQ

The body can be regarded as translating with a linear velocity v ⃗ S \vec{v}_{\mathrm{S}} v S , while rotating with angular velocity ω ⃗ \vec{\omega} ω about an axis passing through S S S

2.1 Rigid Body Velocity: Spatial Velocity (Twist)

  • Spatial Velocity(Twist) : V S = ( ω ⃗ , v ⃗ S ) \mathcal{V} S=\left( \vec{\omega},\vec{v}{\mathrm{S}} \right) VS=(ω ,v S)
    ω ⃗ \vec{\omega} ω - angular velocity; v ⃗ S \vec{v}_{\mathrm{S}} v S - velocity of the body-fixed point currently coincides with S S S

    For any other body-fixed point Q Q Q, its velocity is v ⃗ Q = v ⃗ S + ω ⃗ × R ⃗ S Q \vec{v}{\mathrm{Q}}=\vec{v}{\mathrm{S}}+\vec{\omega}\times \vec{R}_{\mathrm{SQ}} v Q=v S+ω ×R SQ

  • Twist is a 'physical' quantity (just like linear or angular velocity): It can be represented in any chosen reference point S S S

  • A rigid body with V S = ( ω ⃗ , v ⃗ S ) \mathcal{V} S=\left( \vec{\omega},\vec{v}{\mathrm{S}} \right) VS=(ω ,v S) can be 'thought of' as translating at v ⃗ S \vec{v}_{\mathrm{S}} v S while rotating with angular velocity ω ⃗ \vec{\omega} ω about an axis passing through S S S : This is just one way to interpret the motion.

2.2 Spatial Velocity Representation in a Reference Frame

Given frame { O } \left\{ O \right\} {O} and a spatial velocity V ∈ R 6 \mathcal{V} \in \mathbb{R} ^6 V∈R6

Choose o o o (the origin of { O } \left\{ O \right\} {O}) as the reference point to represent the rigid body velocity------ V O = ( ω ⃗ O , v ⃗ O ) \mathcal{V} ^O=\left( \vec{\omega}^O,\vec{v}^O \right) VO=(ω O,v O) : Coordinates for the V \mathcal{V} V in { O } \left\{ O \right\} {O} , by dafault, we assume the origin of the frame is used as the reference point: V O = V O O \mathcal{V} ^O=\mathcal{V} _{\mathrm{O}}^{O} VO=VOO

Example of Twist ,1:

Example of Twist ,2:

2.3 Change Reference Frame for Twist

Given a twist V \mathcal{V} V , let V A \mathcal{V} ^A VA and V B \mathcal{V} ^B VB be their coordinates in frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B} : V A = [ ω ⃗ A v ⃗ A ] , V B = [ ω ⃗ B v ⃗ B ] \mathcal{V} ^A=\left[ \begin{array}{c} \vec{\omega}^A\\ \vec{v}^A\\ \end{array} \right] ,\mathcal{V} ^B=\left[ \begin{array}{c} \vec{\omega}^B\\ \vec{v}^B\\ \end{array} \right] VA=[ω Av A],VB=[ω Bv B]

  • ω ⃗ A = [ Q B A ] ω ⃗ B \vec{\omega}^A=\left[ Q_{\mathrm{B}}^{A} \right] \vec{\omega}^B ω A=[QBA]ω B
  • v ⃗ A A = v ⃗ B A + ω ⃗ A × R ⃗ B A A = [ Q B A ] v ⃗ B B + R ⃗ B A × ( [ Q B A ] ω ⃗ B ) = [ Q B A ] v ⃗ B B + R ⃗ ~ B A [ Q B A ] ω ⃗ B \vec{v}{\mathrm{A}}^{A}=\vec{v}{\mathrm{B}}^{A}+\vec{\omega}^A\times \vec{R}{\mathrm{BA}}^{A}=\left[ Q{\mathrm{B}}^{A} \right] \vec{v}{\mathrm{B}}^{B}+\vec{R}{\mathrm{B}}^{A}\times \left( \left[ Q_{\mathrm{B}}^{A} \right] \vec{\omega}^B \right) =\left[ Q_{\mathrm{B}}^{A} \right] \vec{v}{\mathrm{B}}^{B}+\tilde{\vec{R}}{\mathrm{B}}^{A}\left[ Q_{\mathrm{B}}^{A} \right] \vec{\omega}^B v AA=v BA+ω A×R BAA=[QBA]v BB+R BA×([QBA]ω B)=[QBA]v BB+R ~BA[QBA]ω B

V A = [ ω ⃗ A v ⃗ A ] = [ [ Q B A ] 0 3 × 3 R ⃗ ~ B A [ Q B A ] [ Q B A ] ] 6 × 6 [ ω ⃗ B v ⃗ B ] = [ X B A ] V B \mathcal{V} ^A=\left[ \begin{array}{c} \vec{\omega}^A\\ \vec{v}^A\\ \end{array} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{B}}^{A} \right]& 0_{3\times 3}\\ \tilde{\vec{R}}{\mathrm{B}}^{A}\left[ Q{\mathrm{B}}^{A} \right]& \left[ Q_{\mathrm{B}}^{A} \right]\\ \end{matrix} \right] {6\times 6}\left[ \begin{array}{c} \vec{\omega}^B\\ \vec{v}^B\\ \end{array} \right] =\left[ X{\mathrm{B}}^{A} \right] \mathcal{V} ^B VA=[ω Av A]=[[QBA]R ~BA[QBA]03×3[QBA]]6×6[ω Bv B]=[XBA]VB

If configuration V B \mathcal{V} ^B VB in V A \mathcal{V} ^A VA is [ T B A ] = ( [ Q B A ] , R ⃗ B A ) \left[ T_{\mathrm{B}}^{A} \right] =\left( \left[ Q_{\mathrm{B}}^{A} \right] ,\vec{R}{\mathrm{B}}^{A} \right) [TBA]=([QBA],R BA) , then : [ X B A ] = [ A d T ] = [ [ Q B A ] 0 R ⃗ ~ B A [ Q B A ] [ Q B A ] ] \left[ X{\mathrm{B}}^{A} \right] =\left[ Ad_{\mathrm{T}} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{B}}^{A} \right]& 0\\ \tilde{\vec{R}}{\mathrm{B}}^{A}\left[ Q{\mathrm{B}}^{A} \right]& \left[ Q_{\mathrm{B}}^{A} \right]\\ \end{matrix} \right] [XBA]=[AdT]=[[QBA]R ~BA[QBA]0[QBA]]------adjoint to T

3. Geometric Aspect of Twist: Screw Motion

v ⃗ = ∥ v ⃗ ∥ v ⃗ ^ , ω ⃗ = θ ˙ ω ⃗ ^ \vec{v}=\left\| \vec{v} \right\| \hat{\vec{v}},\vec{\omega}=\dot{\theta}\hat{\vec{\omega}} v =∥v ∥v ^,ω =θ˙ω ^

3.1 Screw Motion : Definition

Screw Motion : Standard/ canonical/kə'nɒnɪk(ə)l/典型 motion for rigid body motion

Rotating about an axis while also translating along the axis

Represented by screw axis { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {R q,s^,h} and rotation speed θ ˙ \dot{\theta} θ˙ (derive the linear speed is h θ ˙ h\dot{\theta} hθ˙)

  • s ^ \hat{s} s^ : unit vector in the direction of the rotatin axis
  • R ⃗ q \vec{R}_q R q : any point on the rotation axis
  • h h h : screw pitch which defines the ratio of the linear velocity along with the screw axis to the angular velocity about the screw axis

Theorem(Chasles) : Every rigid body motion can be realized by a screw motion

3.2 From Screw Motion to Twist

Consider a rigid body under a screw motion with screw axis { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {R q,s^,h} and rotation speed θ ˙ \dot{\theta} θ˙

Fix a reference frame V A \mathcal{V} ^A VA with origin A A A

Find the Twist : V A = ( ω ⃗ A , v ⃗ A A ) = ( s ^ A θ ˙ , v ⃗ q A + ω ⃗ A × R ⃗ q A A ) = ( s ^ A θ ˙ , h θ ˙ + ( s ^ A θ ˙ ) × R ⃗ q A A ) = ( s ^ A θ ˙ , h θ ˙ + R ⃗ q A × ( s ^ A θ ˙ ) ) \mathcal{V} ^A=\left( \vec{\omega}^A,\vec{v}{\mathrm{A}}^{A} \right) =\left( \hat{s}^A\dot{\theta},\vec{v}{\mathrm{q}}^{A}+\vec{\omega}^A\times \vec{R}{\mathrm{qA}}^{A} \right) =\left( \hat{s}^A\dot{\theta},h\dot{\theta}+\left( \hat{s}^A\dot{\theta} \right) \times \vec{R}{\mathrm{qA}}^{A} \right) =\left( \hat{s}^A\dot{\theta},h\dot{\theta}+\vec{R}_{\mathrm{q}}^{A}\times \left( \hat{s}^A\dot{\theta} \right) \right) VA=(ω A,v AA)=(s^Aθ˙,v qA+ω A×R qAA)=(s^Aθ˙,hθ˙+(s^Aθ˙)×R qAA)=(s^Aθ˙,hθ˙+R qA×(s^Aθ˙))

Result: given screw axis { R ⃗ q , s ^ , h } \left\{ \vec{R}q,\hat{s},h \right\} {R q,s^,h} with rotation speed θ ˙ \dot{\theta} θ˙ , the corresponding twist V A = ( ω ⃗ A , v ⃗ A A ) \mathcal{V} ^A=\left( \vec{\omega}^A,\vec{v}{\mathrm{A}}^{A} \right) VA=(ω A,v AA) is given by :
ω ⃗ A = s ^ A θ ˙ , v ⃗ A A = h θ ˙ + R ⃗ q A × ( s ^ A θ ˙ ) \vec{\omega}^A=\hat{s}^A\dot{\theta},\vec{v}{\mathrm{A}}^{A}=h\dot{\theta}+\vec{R}{\mathrm{q}}^{A}\times \left( \hat{s}^A\dot{\theta} \right) ω A=s^Aθ˙,v AA=hθ˙+R qA×(s^Aθ˙)

The result holds as long as all the vectors and the twist are repersented in the same reference frame

3.3 From Twist to Screw Motion

The converse is true as well: given any twist V A = ( ω ⃗ A , v ⃗ A A ) \mathcal{V} ^A=\left( \vec{\omega}^A,\vec{v}_{\mathrm{A}}^{A} \right) VA=(ω A,v AA) we can always find the corresponding screw motion { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {R q,s^,h} and θ ˙ \dot{\theta} θ˙

  • If ω ⃗ = 0 \vec{\omega}=0 ω =0, then it is a pure translation( h = ∞ h=\infty h=∞)
    s ^ = v ⃗ ∥ v ⃗ ∥ , θ ˙ = ∥ v ⃗ ∥ , h = ∞ \hat{s}=\frac{\vec{v}}{\left\| \vec{v} \right\|},\dot{\theta}=\left\| \vec{v} \right\| ,h=\infty s^=∥v ∥v ,θ˙=∥v ∥,h=∞, R ⃗ q \vec{R}_q R q can be arbitrary
  • If ω ⃗ ≠ 0 \vec{\omega}\ne 0 ω =0:
    s ^ = ω ⃗ ∥ ω ⃗ ∥ , θ ˙ = ∥ ω ⃗ ∥ , R ⃗ q = ω ⃗ × v ⃗ ∥ ω ⃗ ∥ 2 , h = ω ⃗ T v ⃗ ∥ ω ⃗ ∥ \hat{s}=\frac{\vec{\omega}}{\left\| \vec{\omega} \right\|},\dot{\theta}=\left\| \vec{\omega} \right\| ,\vec{R}_{\mathrm{q}}=\frac{\vec{\omega}\times \vec{v}}{\left\| \vec{\omega} \right\| ^2},h=\frac{\vec{\omega}^{\mathrm{T}}\vec{v}}{\left\| \vec{\omega} \right\|} s^=∥ω ∥ω ,θ˙=∥ω ∥,R q=∥ω ∥2ω ×v ,h=∥ω ∥ω Tv

You can pluf into the euqation above to very the result

Example: Screw Axin and Twist

3.4 Screw Reoersentation of a Twist

  • Recall : an angular velocity vector ω ⃗ \vec{\omega} ω can be viewed as θ ˙ ω ⃗ \dot{\theta}\vec{\omega} θ˙ω , where ω ⃗ \vec{\omega} ω is the unit ratation axis and θ ˙ \dot{\theta} θ˙ is the rate of rotation about that axis

Similarly, a twist (spatial velocity) V \mathcal{V} V can be interpreted in terms of a screw axis S ^ = { s ^ , h , R ⃗ q } \hat{\mathcal{S}}=\left\{ \hat{s},h,\vec{R}_q \right\} S^={s^,h,R q} and a velocity θ ˙ \dot{\theta} θ˙ about the screw axis

Consider a rigid body motion along a screw axis S ^ = { s ^ , h , R ⃗ q } \hat{\mathcal{S}}=\left\{ \hat{s},h,\vec{R}_q \right\} S^={s^,h,R q} with speed θ ˙ \dot{\theta} θ˙. With slight abuse of notation, we will often write its twist as
V = θ ˙ S ^ \mathcal{V} =\dot{\theta}\hat{\mathcal{S}} V=θ˙S^

In this notation, we think of S ^ \hat{\mathcal{S}} S^ as the twist associated with a unit speed motion along the screw axis { s ^ , h , R ⃗ q } \left\{ \hat{s},h,\vec{R}_q \right\} {s^,h,R q}

4. Extra Note : Tutorial on Twist/spatial Velocity and Screw Axis

4.1 What is Spatial Velocity and Twist

  • Twist/spatial velocity is the velocity of the whole rigid body, not velocity of a particular point
  • Rigid body has inifinitely many points with different velocites
  • All these velocites v ⃗ P i \vec{v}{\mathrm{P}{\mathrm{i}}} v Pi are not independent, depend on the vector of its location and other parameters(common for the entire body), and can be experssed by same set of parameters(twist/spatial velocity is one such parameters)
  1. Assume P 0 P_0 P0 is on the rotation axis/body-fixed, then any other body-fixed
    pt. v ⃗ P i = v ⃗ P 0 + ω ⃗ × R ⃗ P 0 P i \vec{v}{\mathrm{P}{\mathrm{i}}}=\vec{v}_{\mathrm{P}0}+\vec{\omega}\times \vec{R}{\mathrm{P}0\mathrm{P}{\mathrm{i}}} v Pi=v P0+ω ×R P0Pi
  2. What if we use v ⃗ q \vec{v}{\mathrm{q}} v q as the reference velocity, for arbitrary body-fixed
    pt. q q q (may not be on rotation axis), we still have the same expression : v ⃗ P i = v ⃗ q + ω ⃗ × R ⃗ q P i \vec{v}    {\mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{q}}+\vec{\omega}\times \vec{R}{\mathrm{qP}{\mathrm{i}}} v Pi=v q+ω ×R qPi
    ------use P 0 P_0 P0 as intermediate variable , q q q: body-fixed by above------ v ⃗ q = v ⃗ P 0 + ω ⃗ × R ⃗ P 0 P i = v ⃗ P i − ω ⃗ × R ⃗ q P i + ω ⃗ × R ⃗ P 0 P i \vec{v}{\mathrm{q}}=\vec{v}{P_0}+\vec{\omega}\times \vec{R}{P_0\mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{P}{\mathrm{i}}}-\vec{\omega}\times \vec{R}{\mathrm{qP}{\mathrm{i}}}+\vec{\omega}\times \vec{R}{P_0\mathrm{P}{\mathrm{i}}} v q=v P0+ω ×R P0Pi=v Pi−ω ×R qPi+ω ×R P0Pi
  3. Now let;s consider a frame { A } \left\{ A \right\} {A} with origin A A A
    3.1 Assume { A } \left\{ A \right\} {A} body fixed, moves with the body (in this case , let point A A A is point q q q) v ⃗ P i = v ⃗ A + ω ⃗ × R ⃗ A P i \vec{v}{\mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{A}}+\vec{\omega}\times \vec{R}{\mathrm{AP}{\mathrm{i}}} v Pi=v A+ω ×R APi in { A } \left\{ A \right\} {A} system : v ⃗ P i A = v ⃗ A A + ω ⃗ A × R ⃗ A P i A {\vec{v}{\mathrm{P}{\mathrm{i}}}}^A={\vec{v}{\mathrm{A}}}^A+\vec{\omega}^A\times {\vec{R}{\mathrm{AP}{\mathrm{i}}}}^A v PiA=v AA+ω A×R APiA
    3.2 Assume { A } \left\{ A \right\} {A} NOT body-fixed ( { A } \left\{ A \right\} {A} does not move / moves in other way) , let q q q body fixes point such that R ⃗ q = R ⃗ A \vec{R}{\mathrm{q}}=\vec{R}{\mathrm{A}} R q=R A , if we define v ⃗ A \vec{v}{\mathrm{A}} v A as the velocity of the body-fixed point currently coincides wth A A A : v ⃗ P i = v ⃗ q ( t ) + ω ⃗ × R ⃗ q ( t ) P i = v ⃗ A + ω ⃗ × R ⃗ A P i \vec{v}{\mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{q}\left( t \right)}+\vec{\omega}\times \vec{R}{\mathrm{q}\left( t \right) \mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{A}}+\vec{\omega}\times \vec{R}{\mathrm{AP}_{\mathrm{i}}} v Pi=v q(t)+ω ×R q(t)Pi=v A+ω ×R APi

summary : Given twist V = ( ω ⃗ , v ⃗ A ) \mathcal{V} =\left( \vec{\omega}^{},\vec{v}{\mathrm{A}}^{} \right) V=(ω ,v A) , v ⃗ A \vec{v}{\mathrm{A}}^{} v A velocity of body-fixed point currenting cioncides with A A A (reference poiny)

  • For any body-fixed P i P_{\mathrm{i}} Pi: v ⃗ P i = v ⃗ A + ω ⃗ × R ⃗ A P i \vec{v}{\mathrm{P}{\mathrm{i}}}=\vec{v}{\mathrm{A}}+\vec{\omega}\times \vec{R}{\mathrm{AP}{\mathrm{i}}} v Pi=v A+ω ×R APi if A A A is origin of frame { A } \left\{ A \right\} {A} : v ⃗ P i A = v ⃗ A A + ω ⃗ A × R ⃗ A P i A {\vec{v}{\mathrm{P}{\mathrm{i}}}}^A={\vec{v}{\mathrm{A}}}^A+\vec{\omega}^A\times {\vec{R}{\mathrm{AP}{\mathrm{i}}}}^A v PiA=v AA+ω A×R APiA
  • We can think the body is translating at velocity v ⃗ A \vec{v}_{\mathrm{A}} v A , while rotating at velocity ω ⃗ \vec{\omega} ω about axis passing through A A A


4.2 What is Screw Motion and Axis?

Screw motion : combined angular and linear motion

  • motion driven by rotation
  • parameters : { s ^ , h , R ⃗ q } \left\{ \hat{s},h,\vec{R}_q \right\} {s^,h,R q} and θ ˙ \dot{\theta} θ˙: ( s ^ , R ⃗ q ) \left( \hat{s},\vec{R}_q \right) (s^,R q) determeines the rotation axis; h h h linear speed / angular speed (due to thread on the screw - rotation induces linear motion)( h = 0 h=0 h=0-pure rotation, h = ∞ h=\infty h=∞-pure translation)

Over all { s ^ , h , R ⃗ q } \left\{ \hat{s},h,\vec{R}_q \right\} {s^,h,R q} screw axis(rotation axis + pitch); θ ˙ \dot{\theta} θ˙ how fast screw rotates

  1. screw motion is a special rigid-body motion, so its has a twist
    pitch a frame { A } \left\{ A \right\} {A} : ( s ^ , h , R ⃗ q ) + θ ˙ ⇒ [ ω ⃗ A v ⃗ A A ] , ω ⃗ A = θ ˙ s ^ , v ⃗ A A = ( h θ ˙ ) s ^ − ω ⃗ A × R ⃗ q \left( \hat{s},h,\vec{R}q \right) +\dot{\theta}\Rightarrow \left[ \begin{array}{c} \vec{\omega}^A\\ \vec{v}{\mathrm{A}}^{A}\\ \end{array} \right] , \vec{\omega}^A=\dot{\theta}\hat{s},\vec{v}_{\mathrm{A}}^{A}=\left( h\dot{\theta} \right) \hat{s}-\vec{\omega}^A\times \vec{R}_q (s^,h,R q)+θ˙⇒[ω Av AA],ω A=θ˙s^,v AA=(hθ˙)s^−ω A×R q use q q q sa the reference
  2. Ant rigid-body motion can be viewed as screw motion. Given any twist V \mathcal{V} V , we can always find ( s ^ , h , R ⃗ q , θ ˙ ) \left( \hat{s},h,\vec{R}_q,\dot{\theta} \right) (s^,h,R q,θ˙)
  3. We know V = S c r e w T o T w i s t ( s ^ , h , R ⃗ q , 1 ) θ ˙ = θ ˙ S \mathcal{V} =ScrewToTwist\left( \hat{s},h,\vec{R}_q,1 \right) \dot{\theta}=\dot{\theta}\mathcal{S} V=ScrewToTwist(s^,h,R q,1)θ˙=θ˙S while S \mathcal{S} S the twist corresponds to screw motion ( s ^ , h , R ⃗ q ) , θ ˙ = 1 \left( \hat{s},h,\vec{R}_q \right) ,\dot{\theta}=1 (s^,h,R q),θ˙=1
    eg. ω ⃗ = θ ˙ ω ^ \vec{\omega}=\dot{\theta}\hat{\omega} ω =θ˙ω^ , ω ^ \hat{\omega} ω^ : rotation axis or angular velocity when rotates about ω ^ \hat{\omega} ω^ at θ ˙ = 1 \dot{\theta}=1 θ˙=1

Summarize : Given any rigid-body motion V = [ ω ⃗ v ⃗ A ] \mathcal{V} =\left[ \begin{array}{c} \vec{\omega}\\ \vec{v}_{\mathrm{A}}^{}\\ \end{array} \right] V=[ω v A] , with angular direction ω ^ \hat{\omega} ω^ linear motion direction v ^ \hat{v} v^ , knows screw axis direction S ^ \hat{\mathcal{S}} S^------ V = θ ˙ S ^ \mathcal{V} =\dot{\theta}\hat{\mathcal{S}} V=θ˙S^

相关推荐
沫儿笙8 小时前
OTC欧地希焊接机器人智能气阀
机器人
爱吃青菜的大力水手8 小时前
软银收购 ABB 机器人业务的背景与 Physical AI 的关联
人工智能·机器人
Mr.Winter`9 小时前
深度强化学习 | 基于SAC算法的动态避障(ROS C++仿真)
人工智能·深度学习·神经网络·机器人·自动驾驶·ros·具身智能
酌量9 小时前
路径平滑优化详解(二次规划): 数学建模与目标函数推导
经验分享·笔记·学习·机器人·自动驾驶
武子康10 小时前
AI-调查研究-102-具身智能 智能机械臂、自动驾驶与人形机器人的模仿学习、强化学习与多模态融合趋势
人工智能·深度学习·机器学习·ai·机器人·强化学习·具身智能
ARM+FPGA+AI工业主板定制专家1 天前
基于JETSON/RK3588机器人高动态双目视觉系统方案
人工智能·机器学习·fpga开发·机器人·自动驾驶
哈泽尔都1 天前
运动控制教学——5分钟学会PRM算法！
人工智能·单片机·算法·数学建模·贪心算法·机器人·无人机
孤独野指针*P1 天前
MoveIt 机械臂碰撞检测：从环境搭建到障碍物规避
linux·3d·机器人
搬砖的小码农_Sky2 天前
人形机器人：Tesla Optimus的AI集成细节
人工智能·ai·机器人
xwz小王子2 天前
Science Robotics 研究综述：基于学习方法的机器人操作动力学模型
机器人·学习方法
热门推荐
01BongoCat - 跨平台键盘猫动画工具02两千字总结：Codex 国内如何安装和使用的教程，以及如何设置中文回答03智能库存管理的需求预测模型：从业务痛点到落地代码的完整实践04GitHub 镜像站点05UV安装并设置国内源06Linux下V2Ray安装配置指南07GitLab 零基础入门指南：从安装到项目管理全流程082025羊城杯网络安全大赛 wp09Cursor Plan Mode：AI 终于知道先想后做了10Spring Boot 实现微信登录，So Easy !