## 3 Method
### Selection of an Appropriate Reduction Ratio
Using the basic idea of maintaining the knee joint torque's compliance with the motor torque speed curve, we substitute \(\dot{q}_{2t}\) into Equations (2)-(4) to calculate the peak knee joint torque \(\tau_{2t}\) (Nm) at time \(t\):  
\[\tau_{2t}= \begin{cases}\tau_{peak}\dot{i}_{r}&\dot{q}_{2t}\leq\frac{2\pi\omega _{break}}{60i_{r}},\\ \tau_{peak}\dot{i}_{r}-\frac{60\tau_{peak}\dot{i}_{r}^{2}q_{2t}}{2\pi(\omega_{ max}-\omega_{break})}&\dot{q}_{2t}\in[\frac{2\pi\omega_{break}}{60i_{r}}, \frac{2\pi\omega_{break}}{60i_{r}}].\end{cases}  \tag{7}\]  
With the joint constraint of Equation (6), the second row of Equation (1) then becomes  
Figure 3: Torque speed curve of the motor K044100-6Y-4.98 Ams when the voltage supplied to the motor is 120 V.  
\[\left( \begin{array}{ccc}M_{21}&M_{22}&M_{23}\end{array} \right)\left( \begin{array}[] {c}-0.5\\ 1\\ -0.5\end{array} \right)\ddot{q}_{2t}+V_{2}+G_{2}=\tau_{2t}, \tag{8}\]  
where \(V_{2},G_{2}\) represent the second rows of \(V,G\) in state \(X_{t}\), respectively. \(M_{ij}(i=2,j=1,\ 2,\ 3)\) is the \(i\)-th row and \(j\)-th column element of \(M\). Therefore, the maximum angular acceleration \(\ddot{q}_{2t}\) of the knee joint at time \(t\) is  
\[\ddot{q}_{2t}=\frac{\tau_{2t}-V_{2}-G_{2}}{\left( \begin{array}{ccc}M_{21}&M_ {22}&M_{23}\end{array} \right)\left( \begin{array}{c}-0.5\\ 1\\ -0.5\end{array} \right)}. \tag{9}\]  
Using the joint constraint Equation (6), we can get the \(q_{t}\), \(\dot{q}_{t}\) and \(\ddot{q}_{t}\) at time \(t\). Thus, the torque of each joint \(\tau_{t}=(\tau_{1t},\tau_{2t},\tau_{3t})\) then becomes  
\[\tau_{t}=M(q_{t})\ddot{q}_{t}+V(q_{t},\dot{q}_{t})+G(q_{t}). \tag{10}\]  
Obviously, the knee joint torque \(\tau_{2t}\) is equal to the peak torque limited by motor characteristics. By taking the derivative of Equation (5) with respect to \(t\), we can calculate the translational acceleration of COM \(\ddot{C}_{t}\) at time \(t\):  
\[\ddot{C}_{t}=\dot{J}(q_{t})\dot{q}_{t}+J(q_{t})\ddot{q}_{t}. \tag{11}\]  
Integration of \(\ddot{q}_{t}\) with respect to \(t\) allows the next moment state \(X_{t+1}(q_{t+1},\dot{q}_{t+1})\) to be obtained. Therefore, by providing the initial state \(X_{0}(q_{0},q_{0})\), we can then calculate iteratively the \(X_{t}\), \(\dot{q}_{t}\), \(\tau_{t}\), \(C_{t}\), \(\dot{C}_{t}\), and \(\ddot{C}_{t}\) throughout the entire push-off phase, which guarantees that the COM maintains maximum acceleration. The control system is shown in Figure 4.  
Since our objective is to jump, we have to detect the take-off time to make the robot foot free when the robot is going to take off. At the beginning of the motion, the \(\dot{z}_{t}\) is positive and the COM is accelerated. When the leg is going to full extension, the \(\dot{z}_{t}\) will decrease rapidly. Since the computation is iterative and has a certain computation period, the take-off time most likely exists within one of the computation periods. The vertical acceleration of COM calculated is \(\leq-g\) before the take-off time, and \(\leq-g\) after the take-off time. In addition, \(\dot{z}_{t+1}\leq-g\) is a condition that only happens in mathematical calculations not in the real jump motion. If we get a vertical acceleration of COM, which is \(\leq-g\) during the push-off phase, it means that the robot has left the ground. The jump height \(h\) is then estimated as  
Figure 4: The control system for choosing an appropriate reduction radio for motor K044100-6Y-4.98 Ams. \[h=\frac{\dot{z}_{takeoff}^{2}}{2g}, \tag{12}\]  
where \(\dot{z}_{takeoff}\) is the vertical velocity of the COM at the take-off time and \(g\) is the acceleration due to gravity. To select the appropriate reduction ratio, we define the same initial state \(X_{0}=(-75^{\circ},150^{\circ},\)\(-75^{\circ},\)\(0,\)\(0,\)\(0)\) for different values of the reduction ratio \(i_{r}\). The results of the relationship between \(h\) and \(i_{r}\) can then be presented graphically. Referring to Figure 5, it shows that when \(i_{r}=97\), the maximum jump height is approximately \(h_{max}=0.343\) (m).