## 2 Mathematical Modeling Strategy
### Development of Mathematical Model 2  
Considering that the actuators are considered part of the mechanical section of the biped robot, let us consider the model presented in Figure 1. This section develops the geometrical analysis that integrates the mechanical section and the actuator's dynamic movement. Considering the model in Figure 1 and studying the section corresponding to the hip section, the following relations are valid:  
\[\theta_{1}=\theta_{1,1}+\theta_{1,2} \tag{31}\]  
The fixed relation between \((x_{0},y_{0})\) and \((x^{\prime}_{0},y^{\prime}_{0})\) yields  
\[\theta_{1,1}=atan\left(\dfrac{y^{\prime}_{0}-y_{0}}{x^{\prime}_{0}-x_{0}}\right) \tag{32}\]  
Now consider that the length of the selected linear actuator for the hip section \(L^{a}_{1}\) is given by \(L^{a}_{1}=L^{a}_{1,1}+L^{a}_{1,2}\). In view of the mechanical relation between \((x_{0},y_{0})\), \((x^{\prime}_{0},y^{\prime}_{0})\) and \((x_{a},y_{a})\), the following equation is produced:  
\[(L^{a}_{1})^{2}=\left(L^{0}\right)^{2}+\left(L^{f}_{1,1}\right)^{2}-2L^{0}L^{f }_{1,1}cos(\theta_{1}) \tag{33}\]In view of (33), \(\gamma_{1,2}=0.5\pi-\theta_{1,2}\), the following relations are valid:  
\[ \begin{split} cos(\gamma_{1,2})&=\frac{y_{0}-y_{a}}{ L_{1,1}^{f}}=\frac{y_{0}-y_{b}}{L_{1,1}^{f}+L_{1,2}^{f}}=\frac{y_{0}-y_{1}}{L_{1}^{f}} \\ sin(\gamma_{1,2})&=\frac{x_{0}-x_{a}}{L_{1,1}^{f}}= \frac{x_{0}-x_{b}}{L_{1,1}^{f}+L_{1,2}^{f}}=\frac{x_{0}-x_{1}}{L_{1}^{f}}\end{split}  \tag{34}\]  
Using a similar geometrical study for the knee study, the following angular equation is valid:  
\[\theta_{2}=\theta_{2,1}+\theta_{2,2} \tag{35}\]  
The static relation given for \((x_{b},y_{b})\) and \((x_{1},y_{1})\) yields  
\[\theta_{2,1}=atan\left(\frac{y_{b}-y_{1}}{x_{b}-x_{1}}\right) \tag{36}\]  
Taking into account the length of the selected linear actuator for the knee section, \(L_{2}^{a}\) is given by \(L_{2}^{a}=L_{2,1}^{a}+L_{2,2}^{a}\). In view of the mechanical relation between \((x_{b},y_{b})\), \((x_{1},y_{1})\) and \((x_{c},y_{c})\), the following equation is produced:  
\[(L_{2}^{a})^{2}=\left(L_{1,3}^{f}\right)^{2}+\left(L_{2,1}^{f}\right)^{2}-2L_ {1,3}^{f}L_{2,1}^{f}cos(\theta_{2}) \tag{37}\]  
In view of (37), \(\gamma_{2,2}=0.5\pi-\theta_{2,2}\), the following relations are valid:  
\[ \begin{split} cos(\gamma_{2,2})&=\frac{y_{1}-y_{c} }{L_{2,1}^{f}}=\frac{y_{1}-y_{d}}{L_{2,1}^{f}+L_{2,2}^{f}}=\frac{y_{1}-y_{2}}{ L_{2}^{f}}\\ sin(\gamma_{2,2})&=\frac{x_{1}-x_{c}}{L_{2,1}^{f}}= \frac{x_{1}-x_{d}}{L_{2,1}^{f}+L_{2,2}^{f}}=\frac{x_{1}-x_{2}}{L_{2}^{f}}\end{split}  \tag{38}\]  
For the ankle analysis, the corresponding angular equation is valid:  
\[\theta_{3}=\theta_{3,1}+\theta_{3,2} \tag{39}\]  
The static relation given for \((x_{d},y_{d})\) and \((x_{2},y_{2})\) yields  
\[\theta_{3,1}=atan\left(\frac{y_{d}-y_{2}}{x_{d}-x_{2}}\right) \tag{40}\]  
In view of the length of the selected linear actuator for the ankle section, \(L_{3}^{a}\) corresponds to \(L_{3}^{a}=L_{3,1}^{a}+L_{3,2}^{a}\). Considering the mechanical relation between \((x_{d},y_{d})\), \((x_{2},y_{2})\) and \((x_{e},y_{e})\) yields:  
\[(L_{3}^{a})^{2}=\left(L_{2,3}^{f}\right)^{2}+\left(L_{3,1}^{f}\right)^{2}-2L_ {2,3}^{f}L_{3,1}^{f}cos(\theta_{3}) \tag{41}\]  
According to (41), \(\gamma_{3,2}=0.5\pi-\theta_{3,2}\), the following relations are valid:  
\[ \begin{split} cos(\gamma_{3,2})&=\frac{x_{2}-x_{e} }{L_{3,1}^{f}}=\frac{x_{2}-x_{3}}{L_{3}^{f}}\\ sin(\gamma_{3,2})&=\frac{y_{2}-y_{e}}{L_{3,1}^{f}}= \frac{y_{2}-y_{3}}{L_{3}^{f}}\end{split}  \tag{42}\]  
According to the mechanical relations described in this section, the kinetic and potential energies can be obtained in the form presented in the Mathematical Model 1. The calculus is omitted just to avoid unnecessary repetition of similar information. Notice that integrating the actuator as an active element of the mechanical section simplifies the analysis of the actuator effect.