## 2 Overview
### Neutron Extraction
Thus, the first issue is finding the optimal view of the moderator in terms of flux and spectrum. Figure 2 shows an image of integrated intensity of useful neutrons at the deflector exit vs. horizontal angle and horizontal shift of the neutron extraction system along the inside of the monolith. A diagonal line of optimal flux is found because the moderator has a preferred direction in which the neutrons are transmitted. Pairing this with the geometrical restrictions of the monolith insert a preferred window of installation is found. Using this, we chose to position the optical axis of SKADI at the coordinates of \(\alpha_{\text{\tiny{br}}}=-0.5^{\circ}\) and \(x_{\text{\tiny{dnd}}}=7\) mm. That is fairly central in the possible window and close to the central region of the high intensity ridge. The deflector, dogle, and bender were examined in order to evaluate the optimum transport mechanism from the monolith to the bunker wall. Restrictions on the materials were the exclusive use of metal substrates inside the bunker and a maximum m-value of m = 4 to prevent too strong degradation due to the high radiation field inside the bunker [10]. The available length inside the bunker is 9 m, including 3.5 m of the monolith insert. The minimum distance to bridge in vertical direction was 20 cm, as this is the width of the anticipated hadronic shower coming from the target and the moderator that should not propagate along the neutron guide to the sample. By vertically shifting  
Figure 2: The intensity map for the moderator view relating horizontal offset of the extraction system along the inside of the moderator and rotation. A window for physically possible settings is outlined in blue. The ridge of highest intensity corresponds to the ratio between offset and rotation to keep looking at the location of highest flux inside the moderator. The inclination of that ridge is angle = 27.3 \(\times\) offset. the neutron guide by the specified 20 cm or more, this background can be harmlessly absorbed in the bunker wall. This vertical displacement also has the advantage of keeping the instrument straight within its assigned angular wedge and avoid interference with neighbouring instruments. The deflector neutron extraction is the finally chosen solution for the neutron extraction system. The most striking reason for that is given by the performance shown in Figure 3. Looking at the spectra and the detailed data of the deflector explains this behavior. As detailed above, one of the primary objectives of the extraction is to block the direct line of sight. This is most easily achieved by inclining the neutron guide or, as in this case, the upper section of the neutron guide until the full cross section of the neutron guide is covered. For SKADI, this means that 3 cm needs to be covered. The restriction of m = 4 and the lowest wavelength of neutrons to be transported of \(\lambda=3\,\)A leads to the following considerations:  
The critical scattering angle of m = 4 is \(Q_{crit}\) = 0.087 A\({}^{-1}\). Using this to find the angle of inclination for the upper surface of the neutron guide, we need to use  
\[Q_{crit}=\frac{4\pi\sin\theta}{\lambda}\rightarrow\theta=\arcsin\frac{Q_{crit} \lambda}{4\pi}, \tag{1}\]  
which yields \(\theta=0.88^{\circ}\), using \(\lambda=2\,\)A. We chose the lower value for \(\lambda\) for this calculation in order to increase the window of accepted neutrons. Finally, we chose \(\theta=0.9^{\circ}\) as the engineering requirement. This shifts the accepted wavelengths to higher values, which is not a drawback as the targeted wavelength band starts at \(\lambda=3\)A. Because the drop in intensity at the critical edge is centered around \(Q_{crit}=0.087\)A\({}^{-1}\) and being in the area of the highest flux, there may still be significant contributions at higher Q-values. Additionally, this gives a slightly relaxed engineering margin, as now the complete cross section of the neutron guide can be covered by an inclined upper neutron mirror in the guide within 2 m using 2 m \(\times\sin 0.9^{\circ}=3.14\) cm.