HCR - Lecture 5 (Original)

Minimum Distance Point-Surface

It’s easy to know if you are “in” or “out” of a single plane

• Knowing your position ${\bar{x}}_a$ ($a$: avatar) or ${\bar{x}}_h$ ($h$: haptic point), and the formula of the plane: $ax+by+cz+d$ we can calculate:

$$a{x}_{ax}+b{x}_{ay}+c{x}_{az}+d$$

and define as “in” if the result is $<0$ or “out” if the result is $>0$.

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Algorithm

Given $x_h$ haptic point (same as $x_a$ just a different name):

$$x_h=<x_{hx},x_{hy},x_{hz}>$$

the minimum distance from the place $\Sigma$ ($ax+by+cz+d=0)$ can be computed as the distance of $x_h$ from $x_p$ ($p$: proxy) where $x_p$ is defined as the projection of $x_h$ onto $\Sigma$.

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Optimization Problem

We can also see this as an optimization problem:

$$\begin{array}{l}\min ||x_p-x_h||\\ x_p\in \Sigma \end{array}$$

Then:

$$\begin{array}{l}||x_p-x_h||=\sqrt{(x_p-x_h{)}^2}\\ \\ \begin{array}{lc}\Rightarrow \min ||x_p-x_h||& =\min (x_p-x_h{)}^2\\ & |\\ & =\min (x_p-x_h{)}^2\\ & |\\ & =\alpha \cdot \min (x_p-x_h{)}^2,\alpha >0\end{array}\end{array}$$

→ We can reformulate the problem as searching for:

$$\min \frac{1}{2}(x_p-x_h{)}^2$$

Being vectors $x_p$ and $x_h$ we can write:

$$\min \frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T$$

And also we rewrite the constraint: $x_p\in \Sigma$, ($ax+by+cz+d=0$) as:

$$[a,b,c]\cdot x_p+d=0$$

Then the problem becomes:

$$\begin{array}{l}\min \frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T\\ [a,b,c]\cdot x_p+d=0\end{array}$$

Lagrangian Approach:

Using the HCR - Lagrangian Theorem we can rewrite the problem as:

$$Q(x_p,\lambda )=\frac{1}{2}(x_p-x_h)(x_p-x_h{)}^T+{\lambda }^T\cdot \left([a,b,c]\cdot x_p+d\right)$$

and search for its minimum, so:

$$(x_p,\lambda ):\{\begin{array}{l}\frac{\partial Q(x_p,\lambda )}{\partial x_p}=0\\ \frac{\partial Q(x_p,\lambda )}{\partial \lambda }=0\end{array}$$

Then we have:

$$(x_p,\lambda ):\{\begin{array}{l}(x_p-x_h)+{\lambda }^T\cdot [a,b,c]=0\\ [a,b,c]\cdot x_p+d=0\end{array}$$

So, with a bit of substitutions we obtain:

$$x_p=-\frac{[a,b,c]}{||[a,b,c]|{|}^2}\cdot d$$

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Triangular Meshes

We can create a plane with 3 points (~ex.: $T_1$: triangle 1, formed by $x_1,x_2,x_3$)

• So given the triangle $T_1$ we can use the previous formula to calculate the minimum distance $x_h$ from $T_1$.

So I can receive an haptic feedback proportional to the distance $\overline{x_h{x}_p}$

• For a smooth object there are no problem with this approach.

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Spigolar Objects

Notice how for a small displacement of $x_h$ the direction of the Force changes drastically.

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Proxy Algorithm

• IDEA: Remember where you entered from.

1. In case of collision, for the first time @ $t_k=0$ (→ $x_h(0)$) we compute the normal proxy and we take the tangent circle for the proxy $x_p$ given a small radius $ϵ$ (in the figure below $C_p(0)$)

2. Then given the haptic point @ $t_k=1$ → $x_h(1)$ we find which triangle in my triangular mesh is “ACTIVE” The ACTIVE triangle is the one that intersect the segment $\overline{x_h(k)C_p(k-1)}$

Let’s see how this proxy algorithm changes things:

• In the new case the proxy cannot change triangle so easily, so it results in a more smooth experience.

How do we change triangle with the new algorithm?

Let’s take for example the following passages:

As we can see there is a small error for $C_p(1)$ but it’s accepted. And for $C_p(2)$ we have indeed changed triangle, now the second triangle is active.

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Friction Cone

A way to visualize friction

The friction cone depends on the $F_D$ (Force Down-ward) and $\mu$ the friction coefficient.

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Proxy Algorithm with Friction Cones:

If $x_h(k+1)does{n}^{\prime }tescapethefrictionconeof$C_p(k)$ then:

$$C_p(k+1)=C_p(k)$$

• The proxy does not move.

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Static and Kinematic Friction

Each material interaction has 2 friction coefficients one static (${\mu }_s$) and one kinematic (${\mu }_d$), the first is to break the stacity and the second to determine the friction force that slows down the motion.

~ Example: wood on wood: ${\mu }_s=0.25\sim 0.5$, ${\mu }_d=0.2$ glass on glass: ${\mu }_s=0.94$, ${\mu }_d=0.4$ sky on dry snow: ${\mu }_s\simeq 0$, ${\mu }_d=0.04$

When the object is moving and ${\mu }_s>>{\mu }_s$ (usually true) we approximate the static friction coefficient ${\mu }_s$ to 0, so we do not need to calculate it, so ${\Omega }_d={\tan }^{-1}({\mu }_d)$

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