# Scoring of Human Body-Balance Ability on Wobble Board Based on the Geometric Solution
## 3 Equivalent Model
### Forward Kinematics  
Figure 1b represents a definition of variables for an equivalent model, which is expressed as an inverted pendulum on a wobble board. \(\Sigma_{0}\) represents the world coordinate, \(\Sigma_{1}\) represents that for the contact point between the wobble and the ground, and \(\Sigma_{2}\) and \(\Sigma_{3}\) represent that for the surface of the wobble, respectively. \(p_{1}^{0}\) indicates the location of \(\Sigma_{1}\) at the Cartesian coordinate system in terms of \(\Sigma_{0}\), \(p_{2}^{0}\) indicates that of \(\Sigma_{2}\) and \(\Sigma_{3}\) in terms of \(\Sigma_{0}\), and \(p_{3}^{0}\) indicates that of the COM, respectively. The inverted pendulum is the pendulum that has its COM above its pivot point. It is unstable and will fall over without additional support. The inverted pendulum can be used as an equivalent model, when the COM of the pendulum is regarded as the COM of the human body [27].  
Figure 2: An example of detected key points as variables in a segmented part (arm), in order to explain how to calculate the location of the center of mass (COM).  
Consider the inverted pendulum with torque actuation at two joints, \(\theta_{2}\) and \(\theta_{3}\), and there is no torque actuation at the joint of \(\theta_{1}\) for the wobble board. Namely, \(\theta_{1}\) is the free joint. The rotation of \(\theta_{1}\) and \(\theta_{2}\) is the mediolateral (ML) axis and that of \(\theta_{3}\) is the anteroposterior (AP) axis, as shown in Figure 1b. \(l_{1}\) represents the distance from the world coordinate of \(\Sigma_{0}\) to the board coordinate of \(\Sigma_{1}\), which is in contact with the ground. \(l_{2}\) represents the height of the wobble board, and \(l_{3}\) represents that from the plane of the wobble board to the COM. The location of \(p_{1}^{0}\), \(p_{2}^{0}\), and \(p_{3}^{0}\) can be calculated by \(q=\left[\theta_{1},\theta_{2},\theta_{3}\right]^{T}\) and all link lengths of \(l_{1}\), \(l_{2}\), and \(l_{3}\)[28]:  
\[p_{1}^{0} =T_{1}^{0}\left[0;0;0;1\right]^{T}\] \[=\begin{bmatrix}c_{1}&-s_{1}&0&l_{1}\\ s_{1}&c_{1}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}l_{1}-l_{2}s_{1}\\ l_{2}c_{1}\\ 0\\ 1\end{bmatrix},\] \[p_{2}^{0} =T_{1}^{0}T_{2}^{1}\left[0;0;0;1\right]^{T}\] \[=\begin{bmatrix}c_{1}&-s_{1}&0&l_{1}\\ s_{1}&c_{1}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}\begin{bmatrix}c_{2}&-s_{2}&0&0\\ s_{2}&c_{2}&0&l_{2}\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}l_{1}-l_{3}s_{12}-l_{2}s_{1}\\ l_{3}c_{12}+l_{2}c_{1}\\ 0\\ 1\end{bmatrix},\] \[p_{3}^{0} =R_{3}^{2}\,p_{3}^{0^{\prime}}\] \[=\begin{bmatrix}1&0&0&0\\ 0&c_{3}&-s_{3}&0\\ 0&s_{3}&c_{3}&0\\ 0&0&0&1\end{bmatrix}\begin{bmatrix}l_{1}-l_{3}s_{12}-l_{2}s_{1}\\ l_{3}c_{12}+l_{2}c_{2}\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}l_{1}-l_{3}s_{12}-l_{2}s_{1}\\ c_{3}(l_{3}c_{12}+l_{2}c_{1})\\ s_{3}(l_{3}c_{12}+l_{2}c_{1})\\ 1\end{bmatrix},\]  
where \(T_{1}^{i-1}\) for \(\mathrm{i}\geq 0\) gives the relationship between the body frame of \(\sum_{i}\) and the body frame of \(\sum_{i-1}\), \(s_{j}\) and \(c_{j}\) for \(\mathrm{j}>0\) are shorthand for \(\mathrm{sin}\)\(\theta_{j}\) and \(\mathrm{cos}\)\(\theta_{j}\), and \(s_{(j)(j+1)}\) and \(c_{(j)(j+1)}\) are shorthand for \(\mathrm{sin}\)\((\theta_{(j)}+\theta_{(j+1)})\) and \(\mathrm{cos}\)\((\theta_{(j)}+\theta_{(j+1)})\). Now, the location of \(p_{3}^{0}\) can be described by the relationship between three angles and links.