## 2 Description of the Control Object  
To synthesize the control algorithm, the mathematical model of the holonomic motion of the WMR was analyzed. The waveforms of motion parameters were determined and served as the predicted values in the control algorithm. For this purpose, the inverse kinematics problem was solved by determining the angular velocities of the mecanum wheels based on the desired motion path of the selected point of the robot's frame. Based on the analysis of the kinematics of the WMR with mecanum wheels in a moving coordinate system [7], it was possible to determine the robot's kinematics in a fixed XYZ system. The WMR diagram is shown in Figure 1.  
In Figure 1, the following notation is introduced: \(\beta\), robot frame rotation angle; \(\alpha\), angle between instantaneous velocity vector \(\
abla_{\text{S}}\) and the x-axis; \(2\text{s}_{\text{y}}\), robot width, \(\text{s}_{\text{y}}=|\text{O}_{3}\text{P}_{2}|=|\text{P}_{2}\text{O}_{4}|\); \(\text{s}_{\text{x}}\), distance between the characteristic point S (platform center of mass) and the front or rear robot axis, \(\text{s}_{\text{x}}=|\text{S}\text{P}_{1}|=|\text{P}_{2}\text{S}|\); \(\delta\), angle between the roller axis and the mecanum wheel axis (\(\delta=45\)); \(\text{x}_{\text{p}}\text{y}_{\text{p}}\text{z}_{\text{p}}\), moving coordinate system associated with the frame of the mobile robot at point S; and \(\text{xyz}\), stationary coordinate system. A view of the WMR model in the \(\text{x}_{\text{p}}\text{z}_{\text{p}}\) plane of the moving coordinate system is presented in Figure 2.  
Figure 1: Diagram of the wheeled mobile robot. Figure 2: View of the WMR model in the \(\text{x}_{\text{p}}\text{z}_{\text{p}}\) plane. The description of WMR kinematics in the stationary coordinate system was formulated in the form of equations [7]:  
\[ \begin{split}&-\dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta+\sin\beta)+ \ \dot{\mathrm{y}}_{\mathrm{s}}(\cos\beta-\sin\beta)+\ \dot{\mathrm{y}}\big{(}\mathrm{s}_{\mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}+ \ \dot{\mathrm{\varphi}}_{1}(\mathrm{R}+\mathrm{r})=0\\ &\dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta-\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(\cos\beta+\sin\beta)+\ \dot{\mathrm{y}}\big{(}\mathrm{s}_{\mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}- \ \dot{\mathrm{\varphi}}_{2}(\mathrm{R}+\mathrm{r})=0\\ &\dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta-\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(\cos\beta+\sin\beta)-\ \dot{\mathrm{y}}\big{(}\mathrm{s}_{\mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}- \ \dot{\mathrm{\varphi}}_{3}(\mathrm{R}+\mathrm{r})=0\\ &-\dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta+\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(\cos\beta-\sin\beta)-\ \dot{\mathrm{y}}\big{(}\mathrm{s}_{\mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}+ \ \dot{\mathrm{\varphi}}_{4}(\mathrm{R}+\mathrm{r})=0\\ \end{split}  \tag{1}\]  
where \(\dot{\mathrm{x}}_{\mathrm{s}},\dot{\mathrm{y}}_{\mathrm{s}}\) are the projections of the velocity vector of point S on the x,y axes of the stationary coordinate system, \(\beta\) is the angular velocity of the robot frame, R is the wheel radius, r is the roller radius, and \(\ \dot{\mathrm{\varphi}}_{1},\ \dot{\mathrm{\varphi}}_{2},\ \dot{\mathrm{\varphi}}_{3},\ \dot{ \mathrm{\varphi}}_{4}\) are the angular velocities of the mecanum wheels. Using the transforming system of Equation (2), the solution of the inverse kinematics problem was determined using the following:  
\[