## 2 Materials and Methods
### Normal Vector Interpolation  
According to the previous hypothesis, the points in \(\mathbf{Q}\) are nearly on the same tangent plane that is associated with projection points and the change of the normal vector is nearly linear in a small space. Using the projection points and their normal vectors, the normal vector for the point \(\mathbf{X}_{t}=(x_{t},y_{t},z_{t})\) can be calculated using a bi-linear interpolation.  
The interpolation involves eight adjacent cubes as shown in Figure 2. For convenience, the cubes are indexed from 1 to 8. Because of the continuity of the unknown isosurfaces, if the cube contains point data and projection point, these points can be considered on the same tangent plane, and the interpolation can be transformed into two dimensions by projection. For convenience, we are projecting along the maximum component of the normal vector.  
Let \(\mathbf{n}_{1}=(n_{x1},\,n_{y1},\,n_{z1})\) is the normal vector of \(\mathbf{p}_{1}\), and \(\,\mid\,n_{z1}\,\mid\,\geq\,\mid\,n_{y1}\,\mid\,\geq\,\mid\,n_{x1}\,\mid\), then the projection direction is the \(Z\) axis. Search the cube 1 and 5, 2 and 6, 3 and 7, 4 and 8 in Figure 2 to get the point \(\mathbf{p}_{1}\), \(\mathbf{p}_{2}\), \(\mathbf{p}_{3}\), \(\mathbf{p}_{4}\) with the normal vectors \(\mathbf{n}_{1}\), \(\mathbf{n}_{2}\), \(\mathbf{n}_{3}\), \(\mathbf{n}_{4}\), separately. Projecting all points in a cube \((i,j,k)\) and  
Figure 2: Indexes of 8 cubes involved in the normal vector interpolation.  
**P1**, **P2**, **P3**, **P4** along the \(Z\) axis, we get projection points in Figure 3. When the one of \(\left|{{n_{x1}}}\right|\) and \(\left|{{n_{y1}}}\right|\) is maximum, the process is the same with search order 1 and 2, 3 and 4, 5 and 6, 7 and 8, or 1 and 3, 2 and 4, 5 and 7, 6 and 8 to get **p1**, **p2**, **p3**, **p4**, and the projection direction should be \(X\) axis or \(Y\) axis.  
Suppose that the projection points \({\bf{p^{\prime}}}_{1}=(x_{1},y_{1})\), \({\bf{p^{\prime}}}_{2}=(x_{2},y_{2})\), \({\bf{p^{\prime}}}_{3}=(x_{3},y_{3})\), \({\bf{p^{\prime}}}_{4}=(x_{4},y_{4})\). The normal vector of point \({\bf{p}}_{a}=(x_{t},y_{a})\) and \({\bf{p}}_{b}=(x_{t},y_{b})\) in Figure 3 can be calculated by linear interpolation between the projection points of \({\bf{p^{\prime}}}_{1}\), \({\bf{p^{\prime}}}_{2}\) and \({\bf{p^{\prime}}}_{3}\), \({\bf{p^{\prime}}}_{4}\) separately. The \(y\) components of \({\bf{p}}_{a}\) and \({\bf{p}}_{b}\) can be obtained by linear interpolation:  
\[\left\{\begin{array}{l}{y_{a}=\frac{x_{2}-x_{t}}{x_{2}-x_{t}}y_{1}+\frac{x_{t }-x_{1}}{x_{2}-x_{1}}y_{2}}\\ {y_{b}=\frac{x_{2}-x_{1}}{x_{2}-x_{1}}y_{3}+\frac{x_{t}-x_{3}}{x_{4}-x_{3}}y_{4 }}\end{array}\right., \tag{15}\]  
In order to get the correct interpolation results. The direction of the normal vectors \({\bf{n}}_{1}\), \({\bf{n}}_{2}\), \({\bf{n}}_{3}\), \({\bf{n}}_{4}\) must be the same. If the dot product \({\bf{n}}_{1}\cdot{\bf{n}}_{2}<0\), then replace \({\bf{n}}_{2}\) with \(-{\bf{n}}_{2}\). After the normal vectors \({\bf{n}}_{2}\), \({\bf{n}}_{3}\), \({\bf{n}}_{4}\) are consistent with \({\bf{n}}_{1}\), the normal vector \({\bf{n}}_{a}\) can be calculated by interpolation between \({\bf{n}}_{1}\) and \({\bf{n}}_{2}\), and the normal vector \({\bf{n}}_{b}\) can be got by interpolation between \({\bf{n}}_{3}\) and \({\bf{n}}_{4}\):  
\[\left\{\begin{array}{l}{\bf{n}}_{a}=\frac{x_{a}-x_{t}}{x_{t}-x_{t}}{\bf{n}}_{ 1}+\frac{x_{t}-x_{t}}{x_{t}-x_{t}}{\bf{n}}_{2}\\ {\bf{n}}_{b}=\frac{x_{t}-x_{t}}{x_{t}-x_{t}}{\bf{n}}_{3}+\frac{x_{t}-x_{t}}{x_ {t}-x_{t}}{\bf{n}}_{4}\end{array}\right., \tag{16}\]  
Then, \({\bf{n}}_{i}\) can be calculated by linear interpolation between \({\bf{n}}_{a}\) and \({\bf{n}}_{b}\):  
\[{\bf{n}}_{t}=\frac{y_{b}-y_{t}}{y_{b}-y_{a}}{\bf{n}}_{a}+\frac{y_{t}-y_{a}}{y_ {b}-y_{a}}{\bf{n}}_{b}, \tag{17}\]  
Without loss of generality, when the tangent plane is not intersected with the cube, the corresponding projection point can be set to \({\bf{p^{\prime}}}_{w}=(0,0)\), in which \(w=1,2,3,4\).