HCR - Lecture 4

HCR - Triangular Meshes

Approximate any surface to a mathematical model made of triangles. Useful for creating simulations and VR-models.

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• How to create a triangular mesh from a real world object:
  1. Scan the object.
  2. Obtain from the scan a cloud of points.
  3. Using the points from the cloud of points create the triangular mesh.

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• Why is it a triangular mesh?
  • Because 3 is the minimum number of points needed to create a plane.

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• Each triangle has its own normal
  • Calculating the normal of a plane is useful for the calculations of the optimal contact point of the robot.

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• ~Ex.: Cloud of points → Triangular mesh:
  

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HCR - Bounding Box

NOT_SURE_ABOUT_THIS Old notes taken really fast

Bounding Box

Let’s define the shape of our triangles with equations: (Triangles equations are planes equations) #NOT_SURE_ABOUT_THIS Nel disegno ho rappresentato due linee, è sbagliato, in quanto $ax+by+cz+d$ in uno spazio 3D rappresenta un piano e non una linea, per ottenere una linea in uno spazio 3D è necessario intersecare due piani.

We can define being in an object if we are inside the bounding box, then we can say that we are “touching” that object.

→ We are in the bounding box if our current position $<x,y,z>$ respects all the inequalities:

$$a_{i}x+b_{i}y+c_{i}z+d\le 0\forall i$$

(this is not entirely true, for just touch it should be $=0$, and for penetration $<0$ but having it exactly equal to 0 is impossible).

The touching simulation and prevision is really good if the mesh has a lot of points (better approximation), but more points also mean more delay.

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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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HCR - How to select a sample triangle

• OCTATREE method:
  • TODO $<?>$ SEARCH FOR OCTATREE METHOD

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Given 8 points create a 3D object (parallelepiped)

• Creating 6 planes that you can play with.

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