# A Method for Precise Tracking Control of Pneumatic Artificial-Muscle-Driven Exoskeletal Robot
## 2 Pneumatic-Muscle-Driven Upper Limb Exoskeletal Robot System
### Handling of Variable Load by Upper Limb Exoskeleton  
First of all, a body-fixed coordination system is built based on the left and right elbow joints, designating the center point of both elbow joints as its point of origin. While lifting with the Upper Limb Exoskeleton, the angle of lifting varies, and the interactive forces between cargo and robot and the vertical component force of gravity will also change accordingly. Thus, the lifting process is a problem of variable load. Taking the left arm, the rotation angle of the elbow joint during lifting is q, as shown in Figure 7. \(a\) represents the distance from the attachment point of the pneumatic muscle to the elbow joint on the upper arm. \(b\) represents the distance from the attachment point of the pneumatic muscle to the elbow joint on the forearm. \(r\) represents the vertical distance from the elbow to the pneumatic muscle. The main parameters of the upper extremity exoskeleton's system designed in this paper are shown in Table 1: a is the distance between the pneumatic muscle and the upper arm connection point and the elbow joint; b is the distance between the pneumatic muscle and the forearm connection point and the elbow joint; m is the mass of the load being carried; the upper extremity exoskeleton's handling process is represented by rotation around the Z axis, and its moment of inertia regarding the Z axis rotation is ZZ. According to the second law of [PERSON], the following analysis can be conducted regarding the lifted object:  
\[T-mg\cos(q)=mr\ddot{q} \tag{20}\]  
\[T=mr\ddot{q}+mg\cos(q) \tag{21}\]  
\[f=mg\sin(q) \tag{22}\]  
Regarding the Upper Limb Exoskeletal Robot, such an analysis of interaction forces can be conducted, from which conclusions can be drawn that the total pressure of the goods carried can be determined by multiplying the pressure in the center-of-mass coordinate system and the direction vector in the world coordination system. Therefore, the total pressure put on the robotic arm can be calculated as follows:  
\[T^{\prime}=T \begin{pmatrix}-\sin(q)\\ \cos(q)\\ 0\end{pmatrix}  \tag{23}\]  
The frictional force between the goods and the robotic arm is:  
\[f^{\prime}=mg\sin(q) \begin{pmatrix}-\cos(q)\\ -\sin(q)\\ 0\end{pmatrix}  \tag{24}\]  
Therefore, under the world coordination system, the joint force put upon the robotic arm is:  
\[F =T \begin{pmatrix}\sin(q)\\ -\cos(q)\\ 0\end{pmatrix} +mg\sin(q) \begin{pmatrix}-\cos(q)\\ -\sin(q)\\ 0\end{pmatrix}  \tag{25}\] \[= \begin{pmatrix}-mr\ddot{q}\sin(q)\\ mr\ddot{q}\cos(q)-mg\\ 0\end{pmatrix} \]  
The passive torque generated by the load on the Upper Limb Exoskeleton robot can be expressed as Equation (24) in this study. By utilizing the principle of symmetry, it can be deduced that this torque is equally divided between the left and right mechanical \begin{table}
\begin{tabular}{p{42.7 pt} p{42.7 pt} p{42.7 pt} p{42.7 pt} p{42.7 pt}} \hline \hline
**Parameter** & **a (m)** & **b (m)** & **m (kg)** & **ZZ (kg.m2)** \\ \hline Value & 0.165 & 0.21 & 5 & 0.05 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Main parameters of the upper extremity exoskeleton. Figure 7: Analysis of force during lifting process. arms. Thus, the torque experienced specifically by the left arm can be calculated using Equation (25). \[p_{\tau}= \begin{pmatrix}r\cos(q)\\ r\sin(q)\\ 0\end{pmatrix} \times F \tag{26}\]  
\[ \begin{smallmatrix}p_{\tau}\\ l\tau=0.5\times p_{\tau}\end{smallmatrix}  \tag{27}\]  
The torque generated by the load causes a downward movement tendency in the Upper Limb Exoskeleton's mechanical arm. To maintain balance, the pneumatic muscles need to provide equal and opposite torque. The control variable in the system is the air pressure. In order to accomplish the task of assisting in lifting, it is necessary to calculate the corresponding air pressure for the pneumatic muscles at this time. Taking the pneumatic muscle on the left mechanical arm as an example, the length of the pneumatic muscle at this time can be expressed using Equation (27).