四元数求导

原创已于 2022-01-26 10:21:06 修改 · 1k 阅读

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于 2022-01-07 17:09:17 首次发布

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${q}=q_{0}+q_{1} i+q_{2} j+q_{3} k$

$q˙=q˙0+q˙1i+q˙2j+q˙3k\dot{q}=\dot{q}_{0}+\dot{q}_{1} i+\dot{q}_{2} j+\dot{q}_{3} k$

$lim⁡t→0q(t+δt)−q(t)δt\quad\lim _{t \rightarrow 0} \frac{\mathrm{q}(t+\delta t)-\mathrm{q}(t)}{\delta t}$

$=lim⁡t→0δq⊗q(t)−q(t)δt=\lim _{t \rightarrow 0} \frac{\delta \mathbf{q} \otimes \mathbf{q}(t)-\mathbf{q}(t)}{\delta t}$

$=lim⁡t→0(δq−1)⊗q(t)δt=\lim _{t \rightarrow 0} \frac{(\delta \mathbf{q}-1) \otimes \mathbf{q}(t)}{\delta t}$

$=lim⁡t→0[0δθ2]⊗q(t)δt=\lim _{t \rightarrow 0}\frac{\left[\begin{array}{l}0 \\ \frac{\delta \theta}{2}\end{array}\right] \otimes \mathbf{q}(t)}{\delta t}$

$ω=lim⁡t→0δθδt\omega=\lim _{t \rightarrow 0} \frac{\delta \theta}{\delta t}$

$ω=[0ωxωyωz]⊤\omega=\begin{bmatrix} 0 & \omega_{x} & \omega_{y} & \omega_{z}\end{bmatrix}^\top$

$q˙=12[0ω]⊗q=12Ω(ω)q\dot{q}=\frac{1}{2}\left[\begin{array}{c}0 \\ \omega\end{array}\right] \otimes \mathbf{q}=\frac{1}{2} \Omega(\omega) q$

$⌊ω]×=[0−ωzωyωz0−ωx−ωyωx0]\lfloor\omega]_{\times}=\left[\begin{array}{ccc}0 & -\omega_{z} & \omega_{y} \\ \omega_{z} & 0 & -\omega_{x} \\ -\omega_{y} & \omega_{x} & 0\end{array}\right]$

$Ω(ω)=[−⌊ω⌋×ω−ωT0]=[0ωz−ωyωx−ωz0ωxωyωy−ωx0ωz−ωx−ωy−ωz0]\Omega(\omega)=\left[\begin{array}{cc}-\lfloor\omega\rfloor_{\times} & \omega \\ -\omega^{T} & 0\end{array}\right]=\left[\begin{array}{cccc}0 & \omega_{z} & -\omega_{y} & \omega_{x} \\ -\omega_{z} & 0 & \omega_{x} & \omega_{y} \\ \omega_{y} & -\omega_{x} & 0 & \omega_{z} \\ -\omega_{x} & -\omega_{y} & -\omega_{z} & 0\end{array}\right]$