Try using eqref

delta-robot.tex

@@ -1,4 +1,4 @@
-% Created 2025-12-02 Tue 15:00
+% Created 2025-12-02 Tue 15:22
 % Intended LaTeX compiler: pdflatex
 \documentclass[a4paper, 10pt, DIV=12, parskip=full, bibliography=totoc]{scrreprt}
 
@@ -28,7 +28,7 @@ The Delta Robot geometry is defined as shown in Figure \ref{fig:delta_robot_sche
 
 The geometry is fully defined by three parameters:
 \begin{itemize}
-\item \texttt{d}: Cube's size (i.e., the length of the cube edge) \cref{eq:detail_kinematics_cubic_s}
+\item \texttt{d}: Cube's size (i.e., the length of the cube edge)
 \item \texttt{a}: Distance from cube's vertex to top flexible joint
 \item \texttt{L}: Distance between two flexible joints (i.e., the length of the struts)
 \end{itemize}
@@ -93,27 +93,29 @@ Let's initialize a Delta Robot architecture, and plot the obtained geometry (Fig
 \end{figure}
 \chapter{Kinematics: Jacobian Matrix and Mobility}
 
-Jacobian matrix between actuator displacement and top platform displacement.
-
 There are three actuators in the following directions \(\hat{s}_1\), \(\hat{s}_2\) and \(\hat{s}_3\);
 
-\begin{equation}\label{eq:detail_kinematics_cubic_s}
+\begin{equation}\label{eq:delta_robot_unit_vectors}
   \hat{\bm{s}}_1 = \begin{bmatrix}  \frac{-1}{\sqrt{6}}       \\ \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} \end{bmatrix}\quad
   \hat{\bm{s}}_2 = \begin{bmatrix}  \frac{\sqrt{2}}{\sqrt{3}} \\  0                  \\ \frac{1}{\sqrt{3}} \end{bmatrix}\quad
   \hat{\bm{s}}_3 = \begin{bmatrix}  \frac{-1}{\sqrt{6}}       \\ \frac{ 1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} \end{bmatrix}
 \end{equation}
 
-\begin{equation}
+The Jacobian matrix is defined as shown in \eqref{eq:delta_robot_jacobian}.
+
+\begin{equation}\label{eq:delta_robot_jacobian}
   \bm{J} = \begin{bmatrix}
   \hat{\bm{s}}_1^T \\ \hat{\bm{s}}_2^T \\ \hat{\bm{s}}_3^T
   \end{bmatrix}
 \end{equation}
 
-\begin{equation}
+It links the small actuator displacement to the top platform displacement \eqref{eq:delta_robot_inverse_kinematics}.
+
+\begin{equation}\label{eq:delta_robot_inverse_kinematics}
   d\mathcal{L} = J d\mathcal{L}
 \end{equation}
 
-\begin{equation}
+\begin{equation}\label{eq:delta_robot_forward_kinematics}
   d\mathcal{X} = J^{-1} d\mathcal{L}
 \end{equation}