[足式机器人]Part3 机构运动学与动力学分析与建模 Ch00-2(4) 质量刚体的在坐标系下运动

2024-01-10 06:08:52

本文仅供学习使用,总结很多本现有讲述运动学或动力学书籍后的总结,从矢量的角度进行分析,方法比较传统,但更易理解,并且现有的看似抽象方法,两者本质上并无不同。

2024年底本人学位论文发表后方可摘抄
若有帮助请引用
本文参考:
黎 旭,陈 强 洪,甄 文 强 等.惯 性 张 量 平 移 和 旋 转 复 合 变 换 的 一 般 形 式 及 其 应 用[J].工 程 数 学 学 报,2022,39(06):1005-1011.

食用方法
质量点的动量与角动量
刚体的动量与角动量——力与力矩的关系
惯性矩阵的表达与推导——在刚体运动过程中的作用
惯性矩阵在不同坐标系下的表达
务必自己推导全部公式,并理解每个符号的含义


H ? Σ M / O F \vec{H}_{\Sigma _{\mathrm{M}}/\mathrm{O}}^{F} H ΣM?/OF?进一步处理可得: H ? Σ M / O F = ∑ i N m P i ? R ? O P i F × ( ω ? F × R ? O P i F ) = ∑ i N m P i ? R ? O P i F × ( ? R ? O P i F × ω ? F ) = ∑ i N m P i ? R ? ~ O P i F ( ? R ? ~ O P i F ) ω ? F \vec{H}_{\Sigma _{\mathrm{M}}/\mathrm{O}}^{F}=\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \vec{R}_{\mathrm{OP}_{\mathrm{i}}}^{F}\times \left( \vec{\omega}^F\times \vec{R}_{\mathrm{OP}_{\mathrm{i}}}^{F} \right)}=\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \vec{R}_{\mathrm{OP}_{\mathrm{i}}}^{F}\times \left( -\vec{R}_{\mathrm{OP}_{\mathrm{i}}}^{F}\times \vec{\omega}^F \right)}=\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \tilde{\vec{R}}_{\mathrm{OP}_{\mathrm{i}}}^{F}\left( -\tilde{\vec{R}}_{\mathrm{OP}_{\mathrm{i}}}^{F} \right)}\vec{\omega}^F H ΣM?/OF?=iN?mPi???R OPi?F?×(ω F×R OPi?F?)=iN?mPi???R OPi?F?×(?R OPi?F?×ω F)=iN?mPi???R ~OPi?F?(?R ~OPi?F?)ω F。进而得出: ? [ I ] = ∑ i N m P i ? R ? ~ O P i F ( ? R ? ~ O P i F ) \Rightarrow \left[ I \right] =\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \tilde{\vec{R}}_{\mathrm{OP}_{\mathrm{i}}}^{F}\left( -\tilde{\vec{R}}_{\mathrm{OP}_{\mathrm{i}}}^{F} \right)} ?[I]=iN?mPi???R ~OPi?F?(?R ~OPi?F?)

2.2.4 牛顿-欧拉方程 Netwon-Euler equation

刚体动力学中常用:
{ F ? Σ M F = m t o t a l ? a ? G F M ? Σ M / G F = [ I ] Σ M / G F α ? M F + ω ? M F × ( [ I ] Σ M / G F ? ω ? M F ) \begin{cases} \vec{F}_{\Sigma _{\mathrm{M}}}^{F}=m_{\mathrm{total}}\cdot \vec{a}_{\mathrm{G}}^{F}\\ \vec{M}_{\Sigma _{\mathrm{M}}/\mathrm{G}}^{F}=\left[ I \right] _{\Sigma _{\mathrm{M}}/\mathrm{G}}^{F}\vec{\alpha}_{\mathrm{M}}^{F}+\vec{\omega}_{\mathrm{M}}^{F}\times \left( \left[ I \right] _{\Sigma _{\mathrm{M}}/\mathrm{G}}^{F}\cdot \vec{\omega}_{\mathrm{M}}^{F} \right)\\ \end{cases} {F ΣM?F?=mtotal??a GF?M ΣM?/GF?=[I]ΣM?/GF?α MF?+ω MF?×([I]ΣM?/GF??ω MF?)?

2.3 惯性矩阵的转换 Inertia-Matrix Transformation

对于空间中的运动刚体而言,刚体的惯性矩阵一般会根据运动坐标系 { M } ?? \left\{ M \right\} \,\, {M}的基矢量为基底进行计算,而不会直接考虑运动刚体在固定坐标系 { F } ?? \left\{ F \right\} \,\, {F}下的惯性矩阵。此时运动坐标系 { M } ?? \left\{ M \right\} \,\, {M}下计算得出的惯性矩阵记为: [ I ] M \left[ I \right] ^M [I]M。若运动坐标系 { M } ?? \left\{ M \right\} \,\, {M}与固定坐标系 { F } ?? \left\{ F \right\} \,\, {F}的基矢量满足: [ i ? M j ? M k ? M ] = [ Q M F ] T [ I ^ J ^ K ^ ] \left[ \begin{array}{c} \vec{i}^M\\ \vec{j}^M\\ \vec{k}^M\\ \end{array} \right] =\left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}}\left[ \begin{array}{c} \hat{I}\\ \hat{J}\\ \hat{K}\\ \end{array} \right] ?i Mj ?Mk M? ?=[QMF?]T ?I^J^K^? ?,其中 [ Q M F ] T \left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}} [QMF?]T转换矩阵Transition Matrix,为正交矩阵Orthogonal Matrix(满足 [ Q M F ] T = [ Q M F ] ? 1 = [ Q F M ] \left[ Q_{\mathrm{M}}^{F} \right] ^T=\left[ Q_{\mathrm{M}}^{F} \right] ^{-1}=\left[ Q_{\mathrm{F}}^{M} \right] [QMF?]T=[QMF?]?1=[QFM?]), [ Q M F ] \left[ Q_{\mathrm{M}}^{F} \right] [QMF?]又称旋转矩阵Rotation~Matrix
(一个向量乘以一个正交阵,相当于对这个向量进行旋转)。也揭示了该矩阵的两个作用:基底转换(转换矩阵 [ Q M F ] T \left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}} [QMF?]T)与向量旋转(旋转矩阵 [ Q M F ] \left[ Q_{\mathrm{M}}^{F} \right] [QMF?]),则考虑最开始的图有:
在这里插入图片描述
R ? P i F = R ? M F + [ Q M F ] R ? P i M \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{F}=\vec{R}_{\mathrm{M}}^{F}+\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} R Pi?F?=R MF?+[QMF?]R Pi?M?

进而分析惯性矩阵,若 O O O 点与固定坐标系原点 F F F 重合,则有:
[ I ] Σ M F = ∑ i N m P i ? [ ( R ? P i F ) T R ? P i F ? E ? R ? P i F ( R ? P i F ) T ] = ∑ i N m P i ? [ ( R ? M F + [ Q M F ] R ? P i M ) T ( R ? M F + [ Q M F ] R ? P i M ) ? E ? ( R ? M F + [ Q M F ] R ? P i M ) ( R ? M F + [ Q M F ] R ? P i M ) T ] = { m t o t a l ? [ ( R ? M F ) T R ? M F ? E ? R ? M F ( R ? M F ) T ] ? [ I 1 ] Σ M F + [ Q M F ] ( ∑ i N m P i ? [ ( R ? P i M ) T R ? P i M ? E ? R ? P i M ( R ? P i M ) T ] ) [ Q M F ] T + ? [ I 2 ] Σ M F m t o t a l ? [ ( R ? M F ) T ( [ Q M F ] R ? C o M M ) ? E ? R ? M F ( [ Q M F ] R ? C o M M ) T ] ? [ I 3 ] Σ M F + m t o t a l ? [ ( [ Q M F ] R ? C o M M ) T R ? M F ? E ? ( [ Q M F ] R ? C o M M ) ( R ? M F ) T ] ? [ I 4 ] Σ M F = [ I 1 ] Σ M F + [ I 2 ] Σ M F + [ I 3 ] Σ M F + [ I 4 ] Σ M F \begin{split} \left[ I \right] _{\Sigma _{\mathrm{M}}}^{F}&=\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{F} \right) ^T\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{F}\cdot E-\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{F}\left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{F} \right) ^T \right]} \\ &=\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( \vec{R}_{\mathrm{M}}^{F}+\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}}\left( \vec{R}_{\mathrm{M}}^{F}+\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) \cdot E-\left( \vec{R}_{\mathrm{M}}^{F}+\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) \left( \vec{R}_{\mathrm{M}}^{F}+\left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}} \right]} \\ &=\left\{ \begin{array}{c} \begin{array}{c} \underbrace{m_{\mathrm{total}}\cdot \left[ \left( \vec{R}_{\mathrm{M}}^{F} \right) ^{\mathrm{T}}\vec{R}_{\mathrm{M}}^{F}\cdot E-\vec{R}_{\mathrm{M}}^{F}\left( \vec{R}_{\mathrm{M}}^{F} \right) ^{\mathrm{T}} \right] }\\ \left[ I_1 \right] _{\Sigma _{\mathrm{M}}}^{F}\\ \end{array}+\\ \begin{array}{c} \underbrace{\left[ Q_{\mathrm{M}}^{F} \right] \left( \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}}\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\cdot E-\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}} \right]} \right) \left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}}+}\\ \left[ I_2 \right] _{\Sigma _{\mathrm{M}}}^{F}\\ \end{array}\\ \begin{array}{c} \underbrace{m_{\mathrm{total}}\cdot \left[ \left( \vec{R}_{\mathrm{M}}^{F} \right) ^{\mathrm{T}}\left( \left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{CoM}}^{M} \right) \cdot E-\vec{R}_{\mathrm{M}}^{F}\left( \left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{CoM}}^{M} \right) ^{\mathrm{T}} \right] }\\ \left[ I_3 \right] _{\Sigma _{\mathrm{M}}}^{F}\\ \end{array}+\\ \begin{array}{c} \underbrace{m_{\mathrm{total}}\cdot \left[ \left( \left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{CoM}}^{M} \right) ^T\vec{R}_{\mathrm{M}}^{F}\cdot E-\left( \left[ Q_{\mathrm{M}}^{F} \right] \vec{R}_{\mathrm{CoM}}^{M} \right) \left( \vec{R}_{\mathrm{M}}^{F} \right) ^{\mathrm{T}} \right] }\\ \left[ I_4 \right] _{\Sigma _{\mathrm{M}}}^{F}\\ \end{array}\\ \end{array} \right. \\ &=\left[ I_1 \right] _{\Sigma _{\mathrm{M}}}^{F}+\left[ I_2 \right] _{\Sigma _{\mathrm{M}}}^{F}+\left[ I_3 \right] _{\Sigma _{\mathrm{M}}}^{F}+\left[ I_4 \right] _{\Sigma _{\mathrm{M}}}^{F} \end{split} [I]ΣM?F??=iN?mPi???[(R Pi?F?)TR Pi?F??E?R Pi?F?(R Pi?F?)T]=iN?mPi???[(R MF?+[QMF?]R Pi?M?)T(R MF?+[QMF?]R Pi?M?)?E?(R MF?+[QMF?]R Pi?M?)(R MF?+[QMF?]R Pi?M?)T]=? ? ?? mtotal??[(R MF?)TR MF??E?R MF?(R MF?)T]?[I1?]ΣM?F??+ [QMF?](iN?mPi???[(R Pi?M?)TR Pi?M??E?R Pi?M?(R Pi?M?)T])[QMF?]T+?[I2?]ΣM?F?? mtotal??[(R MF?)T([QMF?]R CoMM?)?E?R MF?([QMF?]R CoMM?)T]?[I3?]ΣM?F??+ mtotal??[([QMF?]R CoMM?)TR MF??E?([QMF?]R CoMM?)(R MF?)T]?[I4?]ΣM?F???=[I1?]ΣM?F?+[I2?]ΣM?F?+[I3?]ΣM?F?+[I4?]ΣM?F??

其中, [ I 2 ] Σ M F = [ Q M F ] ( ∑ i N m P i ? [ ( R ? P i M ) T R ? P i M ? E ? R ? P i M ( R ? P i M ) T ] ) [ Q M F ] T = [ Q M F ] [ I ] Σ M M [ Q M F ] T \left[ I_2 \right] _{\Sigma _{\mathrm{M}}}^{F}=\left[ Q_{\mathrm{M}}^{F} \right] \left( \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}}\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\cdot E-\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}} \right]} \right) \left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}}=\left[ Q_{\mathrm{M}}^{F} \right] \left[ I \right] _{\Sigma _{\mathrm{M}}}^{M}\left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}} [I2?]ΣM?F?=[QMF?](iN?mPi???[(R Pi?M?)TR Pi?M??E?R Pi?M?(R Pi?M?)T])[QMF?]T=[QMF?][I]ΣM?M?[QMF?]T,对上式进行讨论:

  • 纯回转: R ? M F = 0 \vec{R}_{\mathrm{M}}^{F}=0 R MF?=0时,化简为:
    [ I ] Σ M F ∣ R ? M F = 0 = [ I 2 ] Σ M F = [ Q M F ] ( ∑ i N m P i ? [ ( R ? P i M ) T R ? P i M ? E ? R ? P i M ( R ? P i M ) T ] ) [ Q M F ] T = [ Q M F ] [ I ] Σ M M [ Q M F ] T \left. \left[ I \right] _{\Sigma _{\mathrm{M}}}^{F} \right|_{\vec{\mathrm{R}}_{\mathrm{M}}^{F}=0}=\left[ I_2 \right] _{\Sigma _{\mathrm{M}}}^{F}=\left[ Q_{\mathrm{M}}^{F} \right] \left( \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}}\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\cdot E-\vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M}\left( \vec{R}_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^{\mathrm{T}} \right]} \right) \left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}}=\left[ Q_{\mathrm{M}}^{F} \right] \left[ I \right] _{\Sigma _{\mathrm{M}}}^{M}\left[ Q_{\mathrm{M}}^{F} \right] ^{\mathrm{T}} [I]ΣM?F? ?R MF?=0?=[I2?]ΣM?F?=[QMF?](iN?mPi???[(R Pi?M?)TR Pi?M??E?R Pi?M?(R Pi?M?)T])[QMF?]T=[QMF?][I]ΣM?M?[QMF?]T
  • 纯移动: R ? M F ≠ 0 \vec{R}_{\mathrm{M}}^{F}\ne 0 R MF?=0 [ Q M F ] = E \left[ Q_{\mathrm{M}}^{F} \right] =E [QMF?]=E时,化简为:
    [ I ] Σ M F ∣ R ? M F ≠ 0 , [ Q M F ] = E = [ I 1 ] Σ M F + [ I ] Σ M M \left. \left[ I \right] _{\Sigma _{\mathrm{M}}}^{F} \right|_{\vec{\mathrm{R}}_{\mathrm{M}}^{F}\ne 0,\left[ Q_{\mathrm{M}}^{F} \right] =\mathrm{E}}=\left[ I_1 \right] _{\Sigma _{\mathrm{M}}}^{F}+\left[ I \right] _{\Sigma _{\mathrm{M}}}^{M} [I]ΣM?F? ?R MF?=0,[QMF?]=E?=[I1?]ΣM?F?+[I]ΣM?M?
    上式也称为惯性矩阵的平行轴定理Parallel Axis Theorem
  • 运动坐标系原点与质心点重合: R ? C o M F = 0 \vec{R}_{\mathrm{CoM}}^{F}=0 R CoMF?=0时,化简为:
    [ I ] F ∣ R ? C o M F = 0 = [ I 1 ] + [ I 2 ] \left. \left[ I \right] ^F \right|_{\vec{R}_{\mathrm{CoM}}^{F}=0}=\left[ I_1 \right] +\left[ I_2 \right] [I]F ?R CoMF?=0?=[I1?]+[I2?]

2.4 惯性矩阵的主轴定理} Principal Axis Theorem

进一步观察惯性矩阵:
[ I ] M = [ ∑ i N m P i ? [ ( y P i M ) 2 + ( z P i M ) 2 ] ? ∑ i N m P i ? x P i M y P i M ? ∑ i N m P i ? ( x P i M z P i M ) ? ∑ i N m P i ? ( y P i M x P i M ) ∑ i N m P i ? [ ( x P i M ) 2 + ( z P i M ) 2 ] ? ∑ i N m P i ? ( y P i M z P i M ) ? ∑ i N m P i ? ( z P i M x P i M ) ? ∑ i N m P i ? ( z P i M y P i M ) ∑ i N m P i ? [ ( x P i M ) 2 + ( y P i M ) 2 ] ] \left[ I \right] ^M=\left[ \begin{matrix} \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( y_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2+\left( z_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2 \right]}& -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot x_{\mathrm{P}_{\mathrm{i}}}^{M}y_{\mathrm{P}_{\mathrm{i}}}^{M}}& -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left( x_{\mathrm{P}_{\mathrm{i}}}^{M}z_{\mathrm{P}_{\mathrm{i}}}^{M} \right)}\\ -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left( y_{\mathrm{P}_{\mathrm{i}}}^{M}x_{\mathrm{P}_{\mathrm{i}}}^{M} \right)}& \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( x_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2+\left( z_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2 \right]}& -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left( y_{\mathrm{P}_{\mathrm{i}}}^{M}z_{\mathrm{P}_{\mathrm{i}}}^{M} \right)}\\ -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left( z_{\mathrm{P}_{\mathrm{i}}}^{M}x_{\mathrm{P}_{\mathrm{i}}}^{M} \right)}& -\sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left( z_{\mathrm{P}_{\mathrm{i}}}^{M}y_{\mathrm{P}_{\mathrm{i}}}^{M} \right)}& \sum_i^N{m_{\mathrm{P}_{\mathrm{i}}}\cdot \left[ \left( x_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2+\left( y_{\mathrm{P}_{\mathrm{i}}}^{M} \right) ^2 \right]}\\ \end{matrix} \right] [I]M= ?iN?mPi???[(yPi?M?)2+(zPi?M?)2]?iN?mPi???(yPi?M?xPi?M?)?iN?mPi???(zPi?M?xPi?M?)??iN?mPi???xPi?M?yPi?M?iN?mPi???[(xPi?M?)2+(zPi?M?)2]?iN?mPi???(zPi?M?yPi?M?)??iN?mPi???(xPi?M?zPi?M?)?iN?mPi???(yPi?M?zPi?M?)iN?mPi???[(xPi?M?)2+(yPi?M?)2]? ?,为对称矩阵Symmetric Matrix(此时默认 M M M 点与 F F F 点重合),则一定能够对角化。

等价于找到另一原点与 M M M 重合的坐标系 B B B ,使得: [ I ] B = [ I x x B 0 0 0 I y y B 0 0 0 I z z B ] \left[ I \right] ^B=\left[ \begin{matrix} I_{\mathrm{xx}}^{B}& 0& 0\\ 0& I_{\mathrm{yy}}^{B}& 0\\ 0& 0& I_{\mathrm{zz}}^{B}\\ \end{matrix} \right] [I]B= ?IxxB?00?0IyyB?0?00IzzB?? ?,根据矩阵对角化Matrix Diagonalizing的原理,结合纯回转推导可得:
[ I ] M = [ Q B M ] [ I ] B [ Q B M ] T \left[ I \right] ^M=\left[ Q_{\mathrm{B}}^{M} \right] \left[ I \right] ^B\left[ Q_{\mathrm{B}}^{M} \right] ^{\mathrm{T}} [I]M=[QBM?][I]B[QBM?]T

其中:

  • [ Q B M ] \left[ Q_{\mathrm{B}}^{M} \right] [QBM?] 满足 [ i ? B j ? B k ? B ] = [ Q B M ] T [ i ? M j ? M k ? M ] \left[ \begin{array}{c} \vec{i}^B\\ \vec{j}^B\\ \vec{k}^B\\ \end{array} \right] =\left[ Q_{\mathrm{B}}^{M} \right] ^{\mathrm{T}}\left[ \begin{array}{c} \vec{i}^M\\ \vec{j}^M\\ \vec{k}^M\\ \end{array} \right] ?i Bj ?Bk B? ?=[QBM?]T ?i Mj ?Mk M? ?
  • ( I x x B , I y y B , I z z B ) \left( I_{\mathrm{xx}}^{B},I_{\mathrm{yy}}^{B},I_{\mathrm{zz}}^{B} \right) (IxxB?,IyyB?,IzzB?) 为矩阵 [ I ] M \left[ I \right] ^M [I]M特征值Eigenvalue
  • [ Q B M ] \left[ Q_{\mathrm{B}}^{M} \right] [QBM?] 为对应于特征值矩阵 [ I ] B \left[ I \right] ^B [I]B特征基Standard Eigenvalue Basis(列向量);

在这里插入图片描述

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文章来源:https://blog.csdn.net/LiongLoure/article/details/135480647
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