## 2 Measurement System Modeling
### Mathematical Model  
\[\left[\begin{array}{cc}|f_{\text{north}}|&|f_{\text{east}}|&|f_{\text{ down}}|\end{array}\right]^{\text{T}}\text{ (superscript T indicates transposition) is the projection of geomagnetic vector into NED system, while \(\left[\begin{array}{cc}B_{\text{X}}&B_{\text{Y}}&B_{\text{Z}}\end{array} \right]^{\text{T}}\) is the projection onto body coordinate system. According to (3), they can be expressed as follows  
\[\left[\begin{array}{c}B_{\text{X}}\\ B_{\text{Y}}\\ B_{\text{Z}}\end{array}\right]=\left[\begin{array}{ccc}\cos\psi\cos\theta& \cos\theta\sin\psi&-\sin\theta\\ \cos\psi\sin\varphi\sin\theta-\cos\varphi\sin\psi&\cos\varphi\cos\psi+\sin\gamma \sin\psi\sin\theta&\cos\theta\sin\varphi\\ \sin\varphi\sin\psi+\cos\varphi\cos\psi\sin\theta&\cos\varphi\sin\psi\sin\theta -\cos\psi\sin\varphi&\cos\varphi\cos\theta\end{array}\right]\left[\begin{array} []{c}|f_{\text{north}}|\\ |f_{\text{east}}|\\ |f_{\text{down}}|\end{array}\right] \tag{3}\]  
Figure 2: Here are figures to describe projectile body based on a model of 155 mm artillery projectile launched by howitzer: (**a**) Description of body coordinate system and (**b**) description of the cross-section.  
From the analysis on (3), the mathematical relationship between roll angle and projection of geomagnetic vector into cross-section of projectile is presented as  
\[B_{\mathrm{Z}}=\sqrt{w^{2}+v^{2}}\sin(\varphi+\varepsilon) \tag{4}\]  
where,  
\[w=|f_{\mathrm{north}}|\sin\psi-|f_{\mathrm{east}}|\cos\psi \tag{5}\]  
\[v=|f_{\mathrm{north}}|\cos\psi\sin\theta+|f_{\mathrm{east}}|\sin\psi\sin \theta+|f_{\mathrm{down}}|\cos\theta \tag{6}\]  
\[\varepsilon=\left\{\begin{array}{l}\arctan\left|\frac{v}{w}\right|v\geq 0,w \geq 0\\ 2\pi-\arctan\left|\frac{v}{w}\right|,v\geq 0,w<0\\ \pi-\arctan\left|\frac{v}{w}\right|,v<0,w\geq 0\\ \pi+\arctan\left|\frac{v}{w}\right|,v<0,w<0\end{array}\right. \tag{7}\]  
In Equation (4), \(\varepsilon\) is the compensation angle, and it is the difference between phase angle of \(B_{\mathrm{Z}}\) and \(\varphi\). Moreover, it is irrelevant with total field intensity \(|f|\) but depends on the \(D\), \(I\) and \(\psi\), \(\theta\).  
When the rotation reverses, \(\varphi=-\varphi\) and \(\varepsilon=\pi-\varepsilon\), Equation (4) is presented as  
\[B_{\mathrm{Z}}=\sqrt{w^{2}+v^{2}}\sin(-\varphi+(\pi-\varepsilon)) \tag{8}\]  
Similarly, one can also get the relationship between roll angle and \(B_{\mathrm{Y}}\) from (3) through the derivation above:  
\[B_{\mathrm{Y}}=\sqrt{{w_{1}}^{2}+{v_{1}}^{2}}\sin(\varphi+\varepsilon_{1}) \tag{9}\]  
When the total field is 500 mGs, declination is 59.263\({}^{\circ}\), inclination is \(-6.8285^{\circ}\), pitch is 18\({}^{\circ}\), yaw is 100\({}^{\circ}\), and rotational speed is 20 Hz, the sinusoidal signals and roll angle are shown in Figure 3.  
Shown in Figure 3, the time difference between O\({}_{1}\) and O\({}_{2}\), as well as time difference between O\({}_{1}\) and O\({}_{3}\), indicates the compensation angle \(\varepsilon(\varepsilon_{1})\), constant difference between the phase of \(B_{\mathrm{Z}}(B_{\mathrm{Y}})\) and the roll angle.  
To make a conclusion, if one manages to extract the phase information of \(B_{\mathrm{Z}}\) (or \(B_{\mathrm{Y}}\)) in (4) (or (9)) and figure out the compensation angle \(\varepsilon\) (or \(\varepsilon_{1}\)) which is independent on the amplitude of \(B_{\mathrm{Z}}\) (\(B_{\mathrm{Y}}\)) or the total intensity \(|f|\), the roll angle \(\varphi\) will be obtained.  
Figure 3: Description of compensation angle.