轮速计噪声模型

\[_{O}\widetilde{\mathbf{u}}(t) = _{O}\mathbf{u}(t) + \mathbf{\eta}^{u}(t)\]

轮速计运动模型

\[_{W} \dot{\mathbf{o}} = _{W}\mathbf{u}\] \[_{W} \dot{\mathbf{o}} = \mathrm{R}_{WB} \cdot \mathrm{R}^{B}_{O} \cdot _{O} \mathbf{u}\]

相邻时刻的轮速测量模型

\[_{W}\mathbf{o}(t + \triangle t) = _{W}\mathbf{o}(t) + _{W}\mathbf{u} (t) \triangle t\]

将 \(\mathrm{R}_{WB}\) 记作 \(\mathrm{R}\)，默认位移 \(\mathbf{o}\) 和速度 \(\mathbf{v}\) 为世界坐标系下的量，则：

\[\mathbf{o}(t + \triangle t) = \mathbf{o}(t) + \mathrm{R}_{k} \cdot \mathrm{R}^{B}_{O} \cdot \mathbf{u} \triangle t\]

考虑噪声模型，有：

\[\mathbf{o}(t + \triangle t) = \mathbf{o}(t) + \mathrm{R}_{k} \cdot \mathrm{R}^{B}_{O} \cdot \left( \tilde{\mathbf{u}}(t) - \mathbf{\eta}^{ud}(t) \right) \triangle t\]

关键帧之间的轮速测量模型

\[\mathbf{o}_{j} = \mathbf{o}_{i} + \sum_{k=i}^{j-1} \mathrm{R}_{k} \cdot \mathrm{R}^{B}_{O} \left( \tilde{\mathbf{u}}_{k} - \mathbf{\eta}^{ud}_{k} \right) \triangle t\]

预积分

将第 \(i\) 时刻的状态分离出来：

\[\mathbf{o}_{j} - \mathbf{o}_{i} = \sum_{k=i}^{j-1} \mathrm{R}_{k} \cdot \mathrm{R}^{B}_{O} \left( \tilde{\mathbf{u}}_{k} - \mathbf{\eta}^{ud}_{k} \right) \triangle t\]

并转换到第 \(i\) 时刻的IMU坐标系，得：

\[\begin{align*} \triangle \mathbf{o}_{ij} &= \mathrm{R}_{i}^{T} \left( \mathbf{o}_{j} - \mathbf{o}_{i} \right) \\ &= \sum_{k=i}^{j-1} \triangle \mathrm{R}_{ik} \cdot \mathrm{R}^{B}_{O} \left( \tilde{\mathbf{u}}_{k} - \mathbf{\eta}^{ud}_{k} \right) \triangle t \end{align*}\] \[{\color{red} \triangle \mathbf{o}_{ij} = \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} }\]

分离噪声

将上述预积分公式中得噪声分离出来：

\[\begin{align*} \triangle \mathbf{o}_{ij} &= \sum_{k=i}^{j-1} \triangle \mathrm{R}_{ik} \cdot \mathrm{R}^{B}_{O} \left( \tilde{\mathbf{u}}_{k} - \mathbf{\eta}^{ud}_{k} \right) \triangle t \\ &= \sum_{k=i}^{j-1} \triangle \tilde{\mathrm{R}}_{ik} \left( \mathbf{I} - \delta\phi_{ik}^{\wedge} \right) \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t - \triangle \tilde{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \mathbf{\eta}_{k}^{ud} \triangle t \\ &= \sum_{k=i}^{j-1} \triangle \tilde{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t + \sum_{k=i}^{j-1} \left[ \triangle \tilde{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \delta \phi_{ik} \triangle t - \triangle \tilde{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \mathbf{\eta}_{k}^{ud} \triangle t \right] \\ &= \triangle \tilde{\mathbf{o}}_{ij} - \delta \mathbf{o}_{ij} \end{align*}\]

其中：

\[\begin{align*} \triangle \tilde{\mathbf{o}}_{ij} &= \sum_{k=i}^{j-1} \triangle \tilde{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t \\ &= \triangle \mathbf{o}_{ij} + \delta \mathbf{o}_{ij} \end{align*}\]

噪声传播

\[\begin{align*} \delta \mathbf{o}_{ij} &= \sum_{k=i}^{j-1} \left[ -\triangle \tilde{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \delta \phi_{ik} \triangle t + \triangle \tilde{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \mathbf{\eta}_{k}^{ud} \triangle t \right] \\ &= \sum_{k=i}^{j-2} \left[ -\triangle \tilde{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \delta \phi_{ik} \triangle t + \triangle \tilde{\mathrm{R}}_{ik} \mathbf{\eta}_{k}^{ud} \mathrm{R}^{B}_{O} \triangle t \right] \\ &- \triangle \tilde{\mathrm{R}}_{ij-1} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{j-1} \right)^{\wedge} \delta \phi_{ij-1} \triangle t + \triangle \tilde{\mathrm{R}}_{ij-1} \mathrm{R}^{B}_{O} \mathbf{\eta}_{j-1}^{ud} \triangle t \\ &= \delta \mathbf{o}_{ij-1} - \triangle \tilde{\mathrm{R}}_{ij-1} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{j-1} \right)^{\wedge} \delta \phi_{ij-1} \triangle t + \triangle \tilde{\mathrm{R}}_{ij-1} \mathrm{R}^{B}_{O} \mathbf{\eta}_{j-1}^{ud} \triangle t \end{align*}\]

令预积分的噪声向量为:

\[\delta \mathbf{\eta}_{ik}^{\triangle} = \left[ \delta \mathbf{\phi}_{ik}, \delta \mathbf{v}_{ik}, \delta \mathbf{p}_{ik}, \delta \mathbf{o}_{ik} \right] \in \mathbb{R}^{12}\]

传感器噪声为：

\[\mathbf{\eta}_{k}^{d} = \left[ \mathbf{\eta}_{k}^{gd}, \mathbf{\eta}_{k}^{ad}, \mathbf{\eta}_{k}^{ud} \right] \in \mathbb{R}^{3 \times 3} = \in \mathbb{R}^{9}\]

则预积分噪声的递推公式为：

\[\delta \mathbf{\eta}_{ij}^{\triangle} = \mathbf{A}_{j-1} \delta \mathbf{\eta}_{ij-1}^{\triangle} + \mathbf{B}_{j-1} \mathbf{\eta}_{j-1}^{d}\]

则：

\[\mathbf{A}_{j-1} = \begin{bmatrix} \triangle \tilde{\mathrm{R}}_{j-1 j}^{T} & 0 & 0 & 0 \\ -\triangle \tilde{\mathrm{R}}_{i j-1} \left( \tilde{\mathbf{a}}_{j-1} - \mathbf{b}_{i}^{a} \right)^{\wedge} \triangle t & \mathbf{I} & 0 & 0 \\ -\frac{1}{2} \triangle \tilde{\mathrm{R}}_{i j-1} \left( \tilde{\mathbf{a}}_{j-1} - \mathbf{b}_{i}^{a} \right)^{\wedge} \triangle t^{2} & \triangle t \mathbf{I} & 0 & 0 \\ -\triangle \tilde{\mathrm{R}}_{ij-1} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{j-1} \right)^{\wedge} \triangle t & 0 & 0 & 0 \end{bmatrix} \in \mathbb{R}^{12 \times 12}\] \[\mathbf{B}_{j-1} = \begin{bmatrix} \mathbf{J}_{r}^{j-1} \triangle t & 0 & 0 \\ 0 & \triangle \tilde{\mathrm{R}}_{i j-1} \triangle t & 0 \\ 0 & \frac{1}{2} \triangle \tilde{\mathrm{R}}_{i j-1} \triangle t^{2} & 0 \\ 0 & 0 & \triangle \tilde{\mathrm{R}}_{i j-1} \mathrm{R}^{B}_{O} \triangle t \end{bmatrix} \in \mathbb{R}^{12 \times 9}\]

协方差矩阵递推公式为：

\[\mathbf{\Sigma}_{ij} = \mathbf{A}_{j-1} \mathbf{\Sigma}_{ij-1} \mathbf{A}_{j-1}^{T} + \mathbf{B}_{j-1} \mathbf{\Sigma}_{\mathbf{\eta}}\mathbf{B}_{j-1}^{T} \in \mathbb{R}^{12 \times 12}\] \[\mathbf{\Sigma}_{\mathbf{\eta}} \in \mathbb{R}^{9 \times 9}\]

偏置更新

\[\triangle \tilde{\mathbf{o}}_{ij} \left( \mathbf{b}_{i}^{g} \right) \simeq \triangle \tilde{\mathbf{o}}_{ij} \left( \bar{\mathbf{b}}_{i}^{g} \right) + \frac{\partial{\bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}_{i}^{g}}} \delta \mathbf{b}_{i}^{g}\]

其中：

\[\bar{\mathbf{o}}_{ij} = \tilde{\mathbf{o}}_{ij} \left( \bar{\mathbf{b}}_{i} \right)\]

当偏置进行更新：

\[\hat{\mathbf{b}}_{i} \leftarrow \bar{\mathbf{b}}_{i} + \delta \mathbf{b}_{i}\]

相应的预积分进行更新：

\[\begin{align*} \tilde{\mathbf{o}}_{ij} (\hat{\mathbf{b}}_{i}) &= \sum_{k=i}^{j-1} \triangle \tilde{\mathrm{R}}_{ik} (\hat{\mathbf{b}}_{i}) \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t \\ &= \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \mathbf{Exp} \left( \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g} \right) \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t \\ &= \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \mathbf{I} + \left( \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g} \right)^{\wedge} \right) \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t \\ &= \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t + \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g} \right)^{\wedge} \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \triangle t \\ &= \triangle \bar{\mathbf{o}}_{ij} + \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g} \right)^{\wedge} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right) \triangle t \\ &= \triangle \bar{\mathbf{o}}_{ij} - \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \left( \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g} \right) \triangle t \\ &= \triangle \bar{\mathbf{o}}_{ij} - \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \triangle t \cdot \delta \mathbf{b}_{i}^{g} \\ &= {\color{red} \triangle \bar{\mathbf{o}}_{ij} + \frac{\partial{\triangle \bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}^{g}}} \delta \mathbf{b}_{i}^{g}} \end{align*}\]

所以：

\[\frac{\partial{\triangle \bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}^{g}}} = - \sum_{k=i}^{j-1} \triangle \bar{\mathrm{R}}_{ik} \left( \mathrm{R}^{B}_{O} \tilde{\mathbf{u}}_{k} \right)^{\wedge} \frac{\partial{\triangle \bar{\mathrm{R}}_{ik}}}{\partial{\mathbf{b}^{g}}} \triangle t\]

预积分残差项

\[\begin{align*} \mathbf{r}_{\triangle \mathbf{o}_{ij}} &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} \\ &- \left( \triangle \tilde{\mathbf{o}}_{ij} \left( \bar{\mathbf{b}}_{i}^{g} \right) + \frac{\partial{\bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}_{i}^{g}}} \delta \mathbf{b}_{i}^{g} \right) \\ &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} \\ &- \left( \triangle \bar{\mathbf{o}}_{ij} + \frac{\partial{\bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}_{i}^{g}}} \delta \mathbf{b}_{i}^{g} \right) \end{align*}\]

雅各比求解

\[\begin{array}{l} \mathrm{R}_{i} \leftarrow \mathrm{R}_{i} \operatorname{Exp}\left(\delta \boldsymbol{\phi}_{i}\right), \quad \mathrm{R}_{j} \leftarrow \mathrm{R}_{j} \operatorname{Exp}\left(\delta \boldsymbol{\phi}_{j}\right) \\[2mm] \mathbf{p}_{i} \leftarrow \mathbf{p}_{i}+\mathrm{R}_{i} \delta \mathbf{p}_{i}, \quad \mathbf{p}_{j} \leftarrow \mathbf{p}_{j}+\mathrm{R}_{j} \delta \mathbf{p}_{j} \\[2mm] \mathbf{v}_{i} \leftarrow \mathbf{v}_{i}+\delta \mathbf{v}_{i}, \quad \mathbf{v}_{j} \leftarrow \mathbf{v}_{j}+\delta \mathbf{v}_{i} \\[2mm] \delta \mathbf{b}_{i}^{g} \leftarrow \delta \mathbf{b}_{i}^{g}+\tilde{\delta} \mathbf{b}_{i}^{g}, \quad \delta \mathbf{b}_{i}^{a} \leftarrow \delta \mathbf{b}_{i}^{a}+\tilde{\delta} \mathbf{b}_{i}^{a} \\ \end{array}\]
  • 对 \(\delta \mathbf{p}_{i}\)
\[\begin{align*} \mathbf{r}_{\triangle \mathbf{o}_{ij}} (\mathbf{p}_{i} + \mathrm{R}_{i} \delta \mathbf{p}_{i}) &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} - \mathrm{R}_{i} \delta \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C \\ &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C - \delta \mathbf{p}_{i} \\ &= \mathbf{r}_{\triangle \mathbf{o}_{ij}} (\mathbf{p}_{i}) + (-\mathbf{I}_{3 \times 1}) \delta \mathbf{p}_{i} \end{align*}\]

所以：

\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\delta \mathbf{p}_{i}}} = -\mathbf{I}_{3 \times 1}\]
  • 对 \(\delta \mathbf{p}_{j}\)
\[\begin{align*} \mathbf{r}_{\triangle \mathbf{o}_{ij}} (\mathbf{p}_{j} + \mathrm{R}_{j} \delta \mathbf{p}_{j}) &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} + \mathrm{R}_{j} \delta \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C \\ &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C + (\mathrm{R}_{i}^{T} \mathrm{R}_{j}) \delta \mathbf{p}_{j} \\ &= \mathbf{r}_{\triangle \mathbf{o}_{ij}} (\mathbf{p}_{i}) + (\mathrm{R}_{i}^{T} \mathrm{R}_{j}) \delta \mathbf{p}_{i} \end{align*}\]

所以：

\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\delta \mathbf{p}_{i}}} = \mathrm{R}_{i}^{T} \mathrm{R}_{j}\]
  • 对 \(\delta \mathbf{v}_{i}\)
\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\delta \mathbf{v}_{i}}} = 0\]
  • 对 \(\delta \mathbf{v}_{j}\)
\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\delta \mathbf{v}_{j}}} = 0\]
  • 对 \(\delta \mathbf{\phi}_{i}\)
\[\begin{align*} \mathbf{r}_{\triangle \mathbf{o}_{ij}} \left( \mathrm{R}_{i} \mathbf{Exp} (\delta \phi_{i}) \right) &= \left[ \mathrm{R}_{i} \mathbf{Exp} (\delta \phi_{i}) \right]^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \left[ \mathrm{R}_{i} \mathbf{Exp} (\delta \phi_{i}) \right]^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C \\[2mm] &= \left( \mathbf{I} - \delta \phi_{i}^{\wedge} \right) \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \left( \mathbf{I} - \delta \phi_{i}^{\wedge} \right) \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C \\[2mm] &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C \\ &+ \left( -\delta \phi_{i}^{\wedge} \right) \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} + \mathrm{R}_{j} \mathbf{t}_{O}^{B} \right) \\[2mm] &= \mathbf{r}_{\triangle \mathbf{o}_{ij}} + \left[ \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} + \mathrm{R}_{j} \mathbf{t}_{O}^{B} \right) \right]^{\wedge} \delta \phi_{i} \end{align*}\]

注：

\[\mathbf{Exp} (\phi)^{T} \simeq (\mathbf{I} + \phi^{\wedge})^{T} = \mathbf{I} + (\phi ^{\wedge})^{T} = \mathbf{I} - \phi^{\wedge}\]
  • 对 \(\delta \mathbf{\phi}_{j}\)
\[\begin{align*} \mathbf{r}_{\triangle \mathbf{o}_{ij}} \left( \mathrm{R}_{j} \mathbf{Exp} (\delta \phi_{j}) \right) &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{Exp} (\delta \phi_{j}) \mathbf{t}_{O}^{B} + C \\[2mm] &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} (\mathbf{I} + \delta \phi_{j}^{\wedge}) \mathbf{t}_{O}^{B} + C \\[2mm] &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C + \mathrm{R}_{i}^{T} \mathrm{R}_{j}\delta \phi_{j}^{\wedge} \mathbf{t}_{O}^{B} \\[2mm] &= \mathrm{R}_{i}^{T} \left( \mathbf{p}_{j} - \mathbf{p}_{i} \right) - \mathbf{t}_{O}^{B} + \mathrm{R}_{i}^{T} \mathrm{R}_{j} \mathbf{t}_{O}^{B} + C - \mathrm{R}_{i}^{T} \mathrm{R}_{j} (\mathbf{t}_{O}^{B})^{\wedge} \delta \phi_{j} \end{align*}\]

所以：

\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\delta \phi_{j}}} = - \mathrm{R}_{i}^{T} \mathrm{R}_{j} (\mathbf{t}_{O}^{B})^{\wedge}\]
  • 对 \(\tilde{\delta} \mathbf{b}_{i}^{g}\)
\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\tilde{\delta} \mathbf{b}_{i}^{g}}} = -\frac{\partial{\bar{\mathbf{o}}_{ij}}}{\partial{\mathbf{b}_{i}^{g}}}\]
  • 对 \(\tilde{\delta} \mathbf{b}_{i}^{a}\)
\[\frac{\partial{\mathbf{r}_{\triangle \mathbf{o}_{ij}}}}{\partial{\tilde{\delta} \mathbf{b}_{i}^{a}}} = 0\]