HCR - Interpolation

NOT_SURE_ABOUT_THIS We consider to have the normal to each point/vertex of the triangular mesh, actually we should have the normal of each TRIANGLE of the mesh, so we would need to interpolate especially when we are near a vertex

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For each triangle in our triangular mesh there is only one normal, to solve thins, we can then make an interpolation:

• So the normal now becomes a smooth function (always depending on the position), before as we can see in the figure we have that the normal is constant for each triangle so we have a non-smooth transition when we pass from one triangle to another.

When we have fewer points then we can take the interpolation: → Where:

$$\begin{array}{l}\overline{u_p}=\alpha \overline{u_1}+\beta \overline{u_2}\\ \to x_p=x_1\text{for}\alpha =1,\beta =0\\ \to x_p=x_2\text{for}\alpha =0,\beta =1\end{array}$$

Considering:

• Distance between $x_1$ and $x_2$ as: $\overline{x_1{x}_2}$
• Distance between $x_1$ and $x_p$ as: $\overline{x_1{x}_p}$
• Distance between $x_p$ and $x_2$ as: $\overline{x_p{x}_2}$

All of them positive (distances cannot be negative)

→ Generally we have:

$$\begin{array}{lcc}\beta =& & \frac{\overline{x_1{x}_p}}{\overline{x_1{x}_2}}\\ \alpha =& 1-\beta =& \frac{\overline{x_2{x}_p}}{\overline{x_1{x}_2}}\end{array}$$

Considering now 3 points (for a triangle): $x_1,x_2,x_3$, with their respective normals: $\overline{u_1},\overline{u_2},\overline{u_3}$

→ The interpolated normal of $x_p$ can be calculated as:

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