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B站:CLEAR_LAB
笔者带更新-运动学
课程主讲教师:
Prof. Wei Zhang
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 vA 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 : 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} RPA 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 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: 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
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
[ 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] iBjBkB T P1BP2BP3B = iAjAkA T P1AP2AP3A ⇒ [QBA]T iAjAkA T P1BP2BP3B = iAjAkA T P1AP2AP3A ⇒ iAjAkA T[QBA] P1BP2BP3B = iAjAkA T P1AP2AP3A ⇒[QBA] P1BP2BP3B = P1AP2AP3A = P′1BP′2BP′3B
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} RBA
For a (free) vector R ⃗ f r e e \vec{R}_{\mathrm{free}} Rfree, its coordinates R ⃗ f r e e A \vec{R}_{free}^{A} RfreeA and R ⃗ f r e e B \vec{R}_{free}^{B} RfreeB 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} RfreeA=[QBA]RfreeB
For a point P P P, its coordinates R ⃗ P A \vec{R}_{\mathrm{P}}^{A} RPA and R ⃗ P B \vec{R}_{\mathrm{P}}^{B} RPB 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} RPA=[QBA]RPB+RBA(一个无聊的小陷阱)
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} RfreeA=[QBA]RfreeB——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} RPA=[QBA]RPB+RBA
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] RPA=[QBA]RPB+RBA⇒[RPA1]=[[QBA]01×3RBA1]4×4[RPB1]
⇒ [ 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]0RBA1]
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 [RPA]=[RPA1]∈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] [RPA]=[TBA][RPB]
对于向量 R ⃗ P 1 P 2 A \vec{R}_{\mathrm{P}_1\mathrm{P}_2}^{A} RP1P2A 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] [RP1P2A]=[RP2A−RP1A]=[RP2A1]−[RP1A1]=[RP2A−RP1A0]=[RP1P2A0]
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}} vP be the velocity of the point P P P
parameterize/pə'ræmɪtə,raɪz/参数化
all of their velocities?The body can be regarded as translating with a linear velocity v ⃗ S \vec{v}_{\mathrm{S}} vS , while rotating with angular velocity ω ⃗ \vec{\omega} ω about an axis passing through S S S
Spatial Velocity(Twist) : V S = ( ω ⃗ , v ⃗ S ) \mathcal{V} _S=\left( \vec{\omega},\vec{v}_{\mathrm{S}} \right) VS=(ω,vS)
ω ⃗ \vec{\omega} ω - angular velocity; v ⃗ S \vec{v}_{\mathrm{S}} vS - 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}} vQ=vS+ω×RSQ
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,vS) can be ‘thought of’ as translating at v ⃗ S \vec{v}_{\mathrm{S}} vS 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.
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=(ωO,vO) : 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
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=[ωAvA],VB=[ωBvB]
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=[ωAvA]=[[QBA]R~BA[QBA]03×3[QBA]]6×6[ωBvB]=[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],RBA) , 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
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^
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\} {Rq,s^,h} and rotation speed θ ˙ \dot{\theta} θ˙ (derive the linear speed is h θ ˙ h\dot{\theta} hθ˙)
Theorem(Chasles) : Every rigid body motion can be realized by a screw motion
Consider a rigid body under a screw motion with screw axis { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {Rq,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=(ωA,vAA)=(s^Aθ˙,vqA+ωA×RqAA)=(s^Aθ˙,hθ˙+(s^Aθ˙)×RqAA)=(s^Aθ˙,hθ˙+RqA×(s^Aθ˙))
Result: given screw axis { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {Rq,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=(ωA,vAA) 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) ωA=s^Aθ˙,vAA=hθ˙+RqA×(s^Aθ˙)
The result holds as long as all the vectors and the twist are repersented in the same reference frame
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=(ωA,vAA) we can always find the corresponding screw motion { R ⃗ q , s ^ , h } \left\{ \vec{R}_q,\hat{s},h \right\} {Rq,s^,h} and θ ˙ \dot{\theta} θ˙
You can pluf into the euqation above to very the result
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,Rq} 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,Rq} 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,Rq}
summary : Given twist V = ( ω ⃗ , v ⃗ A ) \mathcal{V} =\left( \vec{\omega}^{},\vec{v}_{\mathrm{A}}^{} \right) V=(ω,vA) , v ⃗ A \vec{v}_{\mathrm{A}}^{} vA velocity of body-fixed point currenting cioncides with A A A (reference poiny)
Screw motion : combined angular and linear motion
Over all { s ^ , h , R ⃗ q } \left\{ \hat{s},h,\vec{R}_q \right\} {s^,h,Rq} screw axis(rotation axis + pitch); θ ˙ \dot{\theta} θ˙ how fast screw rotates
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=[ωvA<