## 2 Instrument Design
### Position of the Sensor to Obtain Optimal Velocity Resolution  
The position of the sensor within the magnetic chamber is crucial in order to obtain the optimal velocity resolution. We obtain the largest change of hit position with particle velocity when the sensor is perpendicular to the aperture (the sensor is mounted on the chamber's surface next to the aperture), as shown in Figure 1b. In this setup, the hit position \(z\) is linked to the speed of a given particle by:  
\[z=2\frac{mv_{0x}}{qB_{0}}. \tag{4}\]  
This expression is based on the assumption that the aperture is reduced to a point and that the particle velocity \(\mathbf{v}_{0}\) before entering the aperture is parallel to the \(x\) axis. In Section 3, we discuss the influence of the aperture size on the velocity determination. A second advantage of this sensor position, apart from having the highest velocity resolution, is the first-order focusing property which states that all particles with velocity \(\mathbf{v}_{0}\) and within a small solid-angle element are focused onto the same hit position in the sensor plane. We demonstrate this property for a particle with \(\mathbf{v}_{0}=v_{0x}\mathbf{e}_{x}+v_{0z}\mathbf{e}_{z}\) and \(\mathbf{r}_{0}=\mathbf{0}\). According to Equation (1), we obtain:  
\[\begin{cases}v_{x}(t)=v_{0x}\cos(\omega_{0}t)-v_{0z}\sin(\omega_{0}t)\\ v_{z}(t)=v_{0z}\cos(\omega_{0}t)+v_{0x}\sin(\omega_{0}t)\end{cases}, \tag{5}\]  
and after integration:  
\[\begin{cases}x(t)=x_{0}+\frac{v_{0x}}{\omega_{0}}\sin(\omega_{0}t)+\frac{v_{0z }}{\omega_{0}}\cos(\omega_{0}t)\\ z(t)=z_{0}+\frac{v_{0z}}{\omega_{0}}\sin(\omega_{0}t)-\frac{v_{0x}}{\omega_{0}} \cos(\omega_{0}t)\end{cases}. \tag{6}\]  
Figure 1: Basic geometry of MPA. The instrument model has a conical aperture of diameter \(a\) at the entrance of the magnetic chamber (yellow area) with a semi angle of \(2.5^{\circ}\). In Figure 1b, we represent the trajectories of three ions with equal mass, charge and speed but different offsets from the central aperture axis. (**a**): Face view; (**b**): Side view.  
Using \(\cos(\alpha)=v_{0x}/\sqrt{v_{0x}^{2}+v_{0z}^{2}}\) and \(\sin(\alpha)=v_{0z}/\sqrt{v_{0x}^{2}+v_{0z}^{2}}\), we obtain:  
\[\begin{cases}x(t)=-\dfrac{\sqrt{v_{0x}^{2}+v_{0z}^{2}}}{\omega_{0}}\sin(\alpha) +\dfrac{\sqrt{v_{0x}^{2}+v_{0z}^{2}}}{\omega_{0}}\left[\cos(\alpha)\sin(\omega _{0}t)+\sin(\alpha)\cos(\omega_{0}t)\right]\\ z(t)=\dfrac{\sqrt{v_{0x}^{2}+v_{0z}^{2}}}{\omega_{0}}\cos(\alpha)+\dfrac{\sqrt{v _{0x}^{2}+v_{0z}^{2}}}{\omega_{0}}\left[\sin(\alpha)\sin(\omega_{0}t)-\cos( \alpha)\cos(\omega_{0}t)\right]\end{cases}, \tag{7}\]  
leading to  
\[\begin{cases}x(t)=-R\sin(\alpha)+R\sin\left(\omega_{0}t+\alpha\right)\\ z(t)=R\cos(\alpha)-R\cos\left(\omega_{0}t+\alpha\right)\end{cases}, \tag{8}\]  
where we define the curvature radius as  
\[R=\sqrt{v_{0x}^{2}+v_{0z}^{2}}/\omega_{0}. \tag{9}\]  
The hit time \(t_{\text{hit}}\) at which the particle hits the sensor plane fulfills the condition \(x(t_{\text{hit}})=0\). From this condition, we find  
\[t_{\text{hit}}=\dfrac{\pi-2\alpha}{\omega_{0}} \tag{10}\]  
resulting in  
\[z(t_{\text{hit}})=2\,R\,\cos(\alpha)=2\,R\left[1-\dfrac{1}{2}\alpha^{2}+o( \alpha^{4})\right]\sim 2\,R. \tag{11}\]  
Thus, for small incoming angles, the hit position is not affected by the initial \(z\)-component of the particle velocity (\(v_{0z}\)). This approximation is used in the remainder of the article to justify our assumption that all ions enter the instrument along its \(x\)-axis, despite potential (small) oblique entry angles of \(\mathbf{v}_{0}\).