## 4 Modeling of Acoustic Target Tracking
### Elimination Condition for Second Auxiliary Result  
We can eliminate a mathematically correct but unfeasible answer most of the times. If the axis line of pencil crosses the up and the down nappes, then one of the answers belongs to the past and the other one belongs to the future. Sensors cannot hear the sound of a target that is going to be generated in the future. So the points that reside on the intersection point of up nappes in Figure 5 are mathematically correct, but according to our application, they are not feasible answers. We are looking for a condition that the axis of pencil crosses both the up and down nappes in Figure 8b. We will use this condition for eliminating one of the infeasible answers of simultaneous equations in target tracking. Let \(N_{ij}\) and \(N_{ij}\) be normal vectors of two planes \(\pi_{ij}\) and \(\pi_{ij}\) of a pencil as they are shown in Figure 8.a. Vectors \(N_{ij}\) and \(N_{ik}\) are given by equations in Equation (28) that can be extracted from equations of sensing planes in Equation (20):  
Figure 7: A sheaf of planes and their intersection points with sensing cones at two points. \[ \begin{cases}N_{lj}=2(x_{j}-x_{i})i+\ 2(y_{j}-y_{i})j\ -2(t_{j}-t_{i})v^{2}k\\ N_{ik}=2(x_{k}-x_{i})i+\ 2(y_{k}-y_{i})j\ -2(t_{k}-t_{i})v^{2}k\end{cases}  \tag{28}\]  
**Figure 8.** (a) Two planes of a pencil and its axis line. (b) Axis of a pencil and its degree with the axis of cone. Let us now represent the outer product of these two normal vectors as vector \(V_{1}\) in Equation (29). This vector is parallel to the axis line of pencil. \[V_{1}=\ N_{ij}\ \times\ N_{ik} \tag{29}\]  
Equation (30) represents the vector \(V_{1}\):  
\[ \begin{array}{l}V_{1}=(-4(y_{j}-y_{i})(t_{k}-t_{i})v^{2}+4(y_{k}-y_{i})(t_{j} -t_{i})v^{2})i\\ \hskip 14.226378 pt+(-4(t_{j}-t_{i})(x_{k}-x_{i})v^{2}+4(t_{k}-t_{i})(x_{j}-x_{i })v^{2})j\\ \hskip 14.226378 pt+(4(x_{j}-x_{i})(y_{k}-y_{i})-4(x_{k}-x_{i})(y_{j}-y_{i}))k \end{array}  \tag{30}\]  
Figure 8.b shows the relative position of an unbounded right vertical cone of sound propagation with vector \(V_{1}\) that is parallel with the axis line of pencil. Vector \(V_{1}\)can be represented in a simpler form as:  
\[ \begin{array}{l}V_{1}=\ a_{1}.i+\ b_{1}.j+\ c_{1}.k\,\begin{cases}a_{1}=\ ((y_{k}-y_{i})(t_{j}-t_{i})v^{2}-(y_{j}-y_{i})(t_{k}-t_{i})v^{2})\\ b_{1}=\ ((t_{k}-t_{i})(x_{j}-x_{i})v^{2}-(t_{j}-t_{i})(x_{k}-x_{i})v^{2})\\ c_{1}=\ ((x_{k}-x_{i})(y_{j}-y_{i})-(x_{j}-x_{i})(y_{k}-y_{i}))\end{cases}\end{array}  \tag{31}\]  
The axis of a vertical right cone of sound propagation of a target object is a normalized vector as follows:  
\[V_{2}=0i+0j+1k \tag{32}\]  
The angle between two vectors in 3-dimensional space can be computed using their internal product [39]. In Figure 8.b, the angle between vector \(V_{1}\) (representing the axis of pencil) and vector \(V_{2}\) (representing the axis of cone) is shown by \(\alpha\) and \(\cos(\alpha)\) can be computed as follows:  
Figure 8. (a) Two planes of a pencil and its axis line. (b) Axis of a pencil and its degree with the axis of cone. \[\tau=cos(\alpha)=\frac{a_{1}.\,a_{2}+b_{1}.\,b_{2}+c_{1}.\,c_{2}}{|V_{1}|.\,|V_{2}|} =\,\frac{c_{1}}{\sqrt{a_{1}{}^{2}+b_{1}{}^{2}+c_{1}{}^{2}}} \tag{33}\]  
The tangent of the angle between the axis of pencil and the axis of sound propagation cone is given by Equation (34):  
\[tan(\alpha)=\sqrt{\frac{1-cos\,\,(\alpha)^{2}}{cos\,\,(\alpha)^{2}}}=\sqrt{ \frac{1-\tau^{2}}{\tau^{2}}} \tag{34}\]  
Based on Equation (18), the tangent of half of aperture angle of sound propagation cone is equal to the sound propagation speed that is \(V=344\,\,m/s\). Relations in Equation (35) summarize the results of target tracking so far.