## 3 Linear Deviations of the Four Scanner Configurations
### Linear Deviations of the Scanner (a) ab-ab  
(1) Using Figure 2(a1), from Appendix A.1, with Equations (A1) and (A5), the maximum linear deviation in the prisms system (obtained for the \(\varphi=0\) case) is  
\[y_{aP}^{max}=\frac{(e-b\cdot\text{tan}\theta_{1})\cdot\text{tan}\epsilon_{3}+ b\cdot\text{tan}\theta_{2}\cdot\text{tan}\epsilon_{3}^{\prime}}{1-\text{tan} \theta_{2}\cdot\text{tan}\epsilon_{3}^{\prime}}, \tag{9}\]  
where the angles \(\epsilon_{3}\) and \(\epsilon_{3}^{\prime}\) are given by Equation (A6).  
From Figure 2(a1), the maximum linear deviation from the second diopter perpendicular on the O.A. to the screen (also for the \(\varphi=0\) case) is  
\[y_{aS}^{max}=(L-I_{3}Q)\cdot\text{tan}D_{a}^{max}, \tag{10}\]  
therefore, using Equation (A3),  
\[y_{aS}^{max}=\left[L-\frac{[b-(e-b\cdot\text{tan}\theta_{1})\cdot\text{tan} \epsilon_{3}]\cdot\text{tan}\theta_{2}}{1-\text{tan}\theta_{2}\cdot\text{tan} \epsilon_{3}^{\prime}}\right]\cdot\text{tan}D_{a}^{max}, \tag{11}\]where the angular deviation \(D_{a}^{max}\) is given by Equation (2).  
(2) Using Figure 2(a2), from Appendix A.2, with Equations (A2) and (A10), the minimum linear deviation (obtained for the \(\varphi=\pi\) case) is  
\[y_{aP}^{min}=\frac{(e-b\cdot tan\theta_{1})\cdot tan\varepsilon_{3}+b\cdot tan \theta_{2}\cdot tan\varepsilon_{3}^{\prime}}{1+tan\theta_{2}\cdot tan \varepsilon_{3}^{\prime}}, \tag{12}\]  
where the angles \(\varepsilon_{3}\) and \(\varepsilon_{3}^{\prime}\) are given in this case by Equation (A6) as well.  
From Figure 2(a2), the minimum linear deviation from the second diopter perpendicular on the O.A. to the screen (also for the \(\varphi=\pi\) case) is  
\[y_{aS}^{min}=\big{(}L-I_{3}Q^{\prime}\big{)}\cdot tanD_{a}^{min}, \tag{13}\]  
therefore, using Equation (A8) and with the same angles,  
\[y_{aS}^{min}=\bigg{[}L-\frac{[b-(e-b\cdot tan\theta_{1})\cdot tan\varepsilon_ {3}]\cdot tan\theta_{2}}{1+tan\theta_{2}\cdot tan\varepsilon_{3}^{\prime}} \bigg{]}\cdot tanD_{a}^{min}, \tag{14}\]  
where the angular deviation \(D_{a}^{min}\) is given by Equation (2).  
For identical prisms, for which \(n_{1}=n_{2}=n\) and \(\theta_{1}=\theta_{2}=\theta\) (therefore \(k=1\)), the radius of the blind spot/the central disk of the scan patterns (Figure 2(b2)) only comprises the first segment in Equation (A7), which is the I\({}_{3}\)B segment given by Equation (A2). This radius of the blind spot is therefore, from Equation (12),  
\[\Delta_{a}=(e-b\cdot tan\theta)tan\varepsilon_{3}=(e-b\cdot tan\theta)\frac{ n\cdot cos\theta-\sqrt{1-n^{2}\cdot sin^{2}\theta}}{n\cdot sin^{2}\theta+ cos\theta\cdot\sqrt{1-n^{2}\cdot sin^{2}\theta}}. \tag{15}\]  
In the particular case of the small angles approximation, this expression becomes  
\[\Delta_{a}^{lin}=(e-b\theta)\frac{n-\sqrt{1-n^{2}\cdot\theta^{2}}}{n\cdot \theta^{2}+\sqrt{1-n^{2}\cdot\theta^{2}}} \tag{16}\]  
In all such equations, the boundary conditions, for example \(1-n^{2}\cdot sin^{2}\theta>0\) in Equation (15), are equivalent with the conditions to avoid total reflections on the second and fourth diopter of the prisms of the scanners in Figure 2--aspects to be discussed in Section 5.1.