Quartz sync: Jun 26, 2024, 10:17 PM

content/Robotics/Inverse Kinematics.md

@@ -38,12 +38,12 @@ $$
 \dot{q}=\Delta q\ \text{dt}.
 \end{aligned}
 $$
-Now we have an iterative algorithm, but how to calculate it, because $f(q)\in \mathbb{R}^6$ involves translation and rotation. Lie group would help! Consider $\mathcal{X}_d$ and $\mathcal{X}_0$, which are the $SE3$ representation of $x_d$ and $x_0$, respectively, we could do $$
+Now we have an iterative algorithm, but how to calculate it, because $f(q)\in \mathbb{R}^6$ involves translation and rotation. Lie group would help! Consider $\mathcal{X}_d$ and $\mathcal{X}_0$, which are the $SE3$ representation of $x_d$ and $x_0$, respectively, we could do 
+$$
 \begin{aligned}
-\Delta q&=J_{q_0}^{-1}\Delta f\\
-\Delta f&=\text{Log}(\mathcal{X}_d-\mathcal{X}_0),
+\Delta q=J_{q_0}^{-1}\text{Log}(\mathcal{X}_d-\mathcal{X}_0),
 \end{aligned}
-$$ $\Delta f$ is a 6D vector in cartesian space. Note that here $-$ means the right subtraction because the Jacobian is define in the local frame. 
+$$ $\text{Log}(\mathcal{X}_d-\mathcal{X}_0)$ is a 6D vector in cartesian space. Note that here $-$ means the right subtraction because the Jacobian is define in the local frame. 
 
 The idea behind is inverse kinematics use the difference of $\mathcal{X}_d$ and $\mathcal{X}_0$ on the $SE(3)$ manifold as a search direction, then map it back to the generalized coordinates. Because
 $$