## 2 Orientation and Localization Equations
### Position Determination
As will be shown in the next steps, its combination with (12) leads to a compact analytical solution, which is derived below: At a given position, the signal power can be calculated by taking the dot product of the corresponding sensor response \(\vec{V}\) with itself, This must be equal to the dot product of \(\vec{B}(y,z)\) with itself, leading to the following equation:  
\[\vec{V}\cdot\vec{V}=|\vec{V}|^{2}=\vec{B}(y,z)\cdot\vec{B}(y,z)\Rightarrow \quad|\vec{V}|^{2}=\frac{4C^{2}(z^{2}+y^{2})}{z^{2}+y^{4}+(2y^{2}_{0}+2y^{2}) z^{2}+y^{4}_{0}-2y^{2}y^{2}_{0}}. \tag{15}\]The most simple equation is found by solving the formula for \(z^{2}\) and inserting it into \(B_{z}(y,z)=V_{z}\). For this, the two solutions for \(z^{2}\) are obtained with \(P=|\vec{V}|\):  
\[z^{2}(y,P)=\pm 2\sqrt{y^{2}y_{0}^{2}-\frac{C^{2}y_{0}^{2}}{P^{2}}+\frac{C^{4}}{ P^{4}}}-y_{0}^{2}-y^{2}+\frac{2C^{2}}{P^{2}}. \tag{16}\]  
Next, with the positive solution of (16), the equation \(B_{z}(y,z^{2}(y,P))=V_{z}\) can be written as:  
\[\frac{-P^{4}y_{0}^{2}+\hat{D}P^{2}y+C^{2}y}{-CP^{2}y_{0}^{2}+2 CD+2C^{3}}=V_{z}, \quad\text{with}\quad\hat{D}=\sqrt{(P^{4}y^{2}-C^{2}P^{2})y_{0}^{2}+C^{4}}.\]  
Further, \(\hat{D}\) is isolated and squared, which leads to:  
\[\hat{D}^{2}=\frac{((P^{4}y-CP^{2}V_{z})y_{0}^{2}-C^{2}P^{2}y+2C^{3}V_{z})^{2}} {(P^{2}y-2 CV_{z})^{2}}.\]  
This eliminates the square root of \(\hat{D}\). After re-substituting and transforming the equation, the position calculation is reduced to a root-finding problem of a fourth-order polynomial,  
\[ \begin{array}{c}\sum_{n=0}^{4}a_{n}y^{n}=0,\qquad a_{4}=P^{8}y_{0}^{2},\;\;a_ {3}=-4 CP^{6}V_{z}y_{0}^{2},\\ a_{2}=-P^{8}y_{0}^{4}+C^{2}P^{6}y_{0}^{2}+4C^{2}P^{4}V_{z}^{2}y_{0}^{2},\\ a_{1}=2C^{6}V_{z}y_{0}^{4}-2C^{3}P^{4}V_{z}y_{0}^{2},\;\;a_{0}=-C^{2}P^{4}V_{z }^{2}y_{0}^{4}.\end{array}  \tag{17}\]  
The general analytic root expressions of such a polynomial are extremely large. Therefore, it is important to make Equation (3) as simple as possible by placing the global coordinate frame exactly halfway between the transmission lines. In addition, the lines should be placed in the \(xy\)-plane parallel to the \(x\)-axis. Then, with the coefficients \(a_{n}\) given in (17), the four large expressions for \(y\) are greatly reduced and become:  
\[y=\pm\frac{\sqrt{P^{4}y_{0}^{2}+2C^{2}V_{z}\sqrt{V_{z}^{2}-P^{2}}+2C^{2}V_{z} ^{2}-C^{2}P^{2}}}{2P^{2}}\pm\frac{\sqrt{P^{4}y_{0}^{2}-2C^{2}V_{z}\sqrt{V_{z}^ {2}-P^{2}}+2C^{2}V_{z}^{2}-C^{2}P^{2}}}{2P^{2}}+\frac{CV_{z}}{P^{2}} \tag{18}\]  
where \(S\) represents the two square root addends. The same result is obtained for the negative solution of Equation (16). Now, in the next step, four different cases can be distinguished, all of which lead to a certain class of solutions. 1. \(V_{y},V_{z}\
eq 0\):  
**Theorem 1**. : _Two of the \(y\)-solutions in (18) can immediately be excluded, because they are complex._  
Proof. : \(V_{z}\leq|\vec{V}|=P,\;\forall\vec{V}\in\mathbb{R}^{3}\Rightarrow\sqrt{V_{z} ^{2}-P^{2}}\in\mathbb{C}\). With this, the above square root addends represented by \(S\) can be written as:  
\[ \begin{array}{c}S=\pm\sqrt{a+j\,b}\pm\sqrt{a-j\,b},\\ a=P^{4}y_{0}^{2}+(2V_{z}^{2}-P^{2})C^{2},\;b=2C^{2}V_{z}\sqrt{P^{2}-V_{z}^{2}}.\end{array}  \tag{19}\]  
Next, the parameters \(r\) and \(\varphi\) are introduced as:  
\[r=\sqrt{a^{2}+b^{2}},\;\;\varphi=tan^{-1}\left(\frac{b}{a}\right)+\pi. \tag{20}\]  
Now, if the addends have a different sign \(S\), this results in:  
\[ \begin{array}{c}S=\pm\sqrt{a+j\,b}\mp\sqrt{a-j\,b}=\pm\sqrt{r}\,e^{j\frac{ \varphi}{2}}\mp\sqrt{r}\,e^{-j\frac{\varphi}{2}}\\ =\pm 2j\sqrt{r}\,\frac{1}{2}(e^{j\frac{\varphi}{2}}-e^{-j\frac{\varphi}{2}}) =\pm 2j\sqrt{r}\,sin\frac{\varphi}{2},\end{array} \]whereby the \(e\)-expressions are obtained using [PERSON]'s formula. Applying this procedure for the case of the same sign, the real-numbered and more compact expressions for \(y\) can be derived.