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

本文主要是介绍[足式机器人]Part4 南科大高等机器人控制课 Ch02 Rigid Body Configuration and Velocity,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

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
    • 1.1 Special Orthogonal Group
    • 1.2 Use of Rotation Matrix
    • 1.3 Homogeneous Transformation Matrix
  • 2. Rigid Body Velocity(Twist)
    • 2.1 Rigid Body Velocity: Spatial Velocity (Twist)
    • 2.2 Spatial Velocity Representation in a Reference Frame
    • 2.3 Change Reference Frame for Twist
  • 3. Geometric Aspect of Twist: Screw Motion
    • 3.1 Screw Motion : Definition
    • 3.2 From Screw Motion to Twist
    • 3.3 From Twist to Screw Motion
    • 3.4 Screw Reoersentation of a Twist
  • 4. Extra Note : Tutorial on Twist/spatial Velocity and Screw Axis
    • 4.1 What is Spatial Velocity and Twist
    • 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 a2b3a3b2a3b1a1b3a1b2a2b1 = I^J^K^ T 0a3a2a30a1a2a10 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 aG,bGabG
  • 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 (ab)c=a(bc),a,b,cG
  • Identity element: ∃ e ∈ G \exists e\in G eG such that e ∙ a = a e\bullet a=a ea=a , for all a ∈ G a\in G aG
  • Inverse element: For each a ∈ G a\in G aG, there is a b ∈ G b\in G bG such that a ∙ b = b ∙ a = e a\bullet b=b\bullet a=e ab=ba=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 pF=[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 = P1BP2BP3B

  • 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 PR3, 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 P2AR P1A]=[R P2A1][R P1A1]=[R P2AR P1A0]=[R P1P2A0]

  • Example of Homogeneous Transformation Matrix
    在这里插入图片描述

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 PSv Q=v Sω ×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 VR6

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:
在这里插入图片描述

Example of Twist ,2:
在这里插入图片描述

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
在这里插入图片描述
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
在这里插入图片描述
在这里插入图片描述

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 Piω ×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

在这里插入图片描述
在这里插入图片描述

4.2 What is Screw Motion and Axis?

在这里插入图片描述
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^

这篇关于[足式机器人]Part4 南科大高等机器人控制课 Ch02 Rigid Body Configuration and Velocity的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/466437

相关文章

Spring Security 基于表达式的权限控制

前言 spring security 3.0已经可以使用spring el表达式来控制授权,允许在表达式中使用复杂的布尔逻辑来控制访问的权限。 常见的表达式 Spring Security可用表达式对象的基类是SecurityExpressionRoot。 表达式描述hasRole([role])用户拥有制定的角色时返回true (Spring security默认会带有ROLE_前缀),去

控制反转 的种类

之前对控制反转的定义和解释都不是很清晰。最近翻书发现在《Pro Spring 5》(免费电子版在文章最后)有一段非常不错的解释。记录一下,有道翻译贴出来方便查看。如有请直接跳过中文,看后面的原文。 控制反转的类型 控制反转的类型您可能想知道为什么有两种类型的IoC,以及为什么这些类型被进一步划分为不同的实现。这个问题似乎没有明确的答案;当然,不同的类型提供了一定程度的灵活性,但

深入解析秒杀业务中的核心问题 —— 从并发控制到事务管理

深入解析秒杀业务中的核心问题 —— 从并发控制到事务管理 秒杀系统是应对高并发、高压力下的典型业务场景,涉及到并发控制、库存管理、事务管理等多个关键技术点。本文将深入剖析秒杀商品业务中常见的几个核心问题,包括 AOP 事务管理、同步锁机制、乐观锁、CAS 操作,以及用户限购策略。通过这些技术的结合,确保秒杀系统在高并发场景下的稳定性和一致性。 1. AOP 代理对象与事务管理 在秒杀商品

基于树梅派的视频监控机器人Verybot

最近这段时间做了一个基于树梅派 ( raspberry pi ) 的视频监控机器人平台 Verybot ,现在打算把这个机器人的一些图片、视频、设计思路进行公开,并且希望跟大家一起研究相关的各种问题,下面是两张机器人的照片:         图片1:                   图片2                    这个平台的基本组成是:

PostgreSQL中的多版本并发控制(MVCC)深入解析

引言 PostgreSQL作为一款强大的开源关系数据库管理系统,以其高性能、高可靠性和丰富的功能特性而广受欢迎。在并发控制方面,PostgreSQL采用了多版本并发控制(MVCC)机制,该机制为数据库提供了高效的数据访问和更新能力,同时保证了数据的一致性和隔离性。本文将深入解析PostgreSQL中的MVCC功能,探讨其工作原理、使用场景,并通过具体SQL示例来展示其在实际应用中的表现。 一、

【高等代数笔记】线性空间(一到四)

3. 线性空间 令 K n : = { ( a 1 , a 2 , . . . , a n ) ∣ a i ∈ K , i = 1 , 2 , . . . , n } \textbf{K}^{n}:=\{(a_{1},a_{2},...,a_{n})|a_{i}\in\textbf{K},i=1,2,...,n\} Kn:={(a1​,a2​,...,an​)∣ai​∈K,i=1,2,...,n

vue2实践:el-table实现由用户自己控制行数的动态表格

需求 项目中需要提供一个动态表单,如图: 当我点击添加时,便添加一行;点击右边的删除时,便删除这一行。 至少要有一行数据,但是没有上限。 思路 这种每一行的数据固定,但是不定行数的,很容易想到使用el-table来实现,它可以循环读取:data所绑定的数组,来生成行数据,不同的是: 1、table里面的每一个cell,需要放置一个input来支持用户编辑。 2、最后一列放置两个b

【电机控制】数字滤波算法(持续更新)

文章目录 前言1. 数字低通滤波 前言 各种数字滤波原理,离散化公式及代码。 1. 数字低通滤波 滤波器公式 一阶低通滤波器的输出 y [ n ] y[n] y[n] 可以通过以下公式计算得到: y [ n ] = α x [ n ] + ( 1 − α ) y [ n − 1 ] y[n] = \alpha x[n] + (1 - \alpha) y[n-1]

OpenStack离线Train版安装系列—3控制节点-Keystone认证服务组件

本系列文章包含从OpenStack离线源制作到完成OpenStack安装的全部过程。 在本系列教程中使用的OpenStack的安装版本为第20个版本Train(简称T版本),2020年5月13日,OpenStack社区发布了第21个版本Ussuri(简称U版本)。 OpenStack部署系列文章 OpenStack Victoria版 安装部署系列教程 OpenStack Ussuri版

OpenStack离线Train版安装系列—1控制节点-环境准备

本系列文章包含从OpenStack离线源制作到完成OpenStack安装的全部过程。 在本系列教程中使用的OpenStack的安装版本为第20个版本Train(简称T版本),2020年5月13日,OpenStack社区发布了第21个版本Ussuri(简称U版本)。 OpenStack部署系列文章 OpenStack Victoria版 安装部署系列教程 OpenStack Ussuri版