HCR - Lecture 4 (Original)

Triangular Meshes

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

→ 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

→ Why is it a triangular mesh?

• Because 3 is the minimum number of points needed to create a plane.

→ 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.

~Ex.: Cloud of points → Triangular mesh

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Bounding Box

Let’s define the shape of our triangles with equations: (Triangles equations are planes equations)

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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Interpolation

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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How to select a sample triangle?

• OCTATREE method.

Given 8 points create a 3D object (parallelepiped) → Creating 6 planes that you can play with.

$<?>$ SEARCH FOR OCTATREE METHOD