HCR - Lecture 5

HCR - 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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HCR - Algorithm for Minimum Distance from Surface

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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HCR - 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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HCR - Proxy Point for 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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HCR - Proxy Algorithm

• Edgy object (Oggetto spigoloso):
  
  • 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.

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• 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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HCR - 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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• Friction Cone:
  
  
  • Where:$\begin{array}{l}\Theta =\\ \eta :\text{friction coefficent}\end{array}$(The friction coefficent ($\eta$) is specific for each pair of object interaction)
  • A force that want to break the friction is applied as shown:
    
    
    • $F_1$ is too small to break friction (the object doesn’t move).
    • $F_2$ is big enough (the object begins to move).
    • Graphically $F_1$ is inside the friction cone.
      While $F_2$ is outside the friction cone
  • Knowing where the friction cone is, and how big it is, is extremely crucial, because if i want to apply force to the object without it slipping out (break friction) i need to apply all the forces such that they stay inside the friction cones.

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• ~Ex.: Position of the friction cones
  • Suppose we have this object:
    
  • Let’s see how the position of the friction cones is useful.
    I apply to the previous object 3 forces: $F_1,F_2,F_3$, resulting in this 3 friction cones:
    
  • Take now an opposite force $F$ that tries to break friction.
    It is clear that where i apply the previous 3 forces changes the intensity of the Force $F$ needed to break friction:
    

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HCR - Proxy Algorithm with Friction Cones

If $x_h(k+1)$ doesn’t escape the friction cone of $C_p(k)$ then:$C_p(k+1)=C_p(k)$Meaning that the proxy does not move.

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HCR - Static and Kinematic Friction

Each material interaction has 2 friction coefficients:

• One static (${\mu }_s$).
• and one kinematic (${\mu }_d$),

The static friction coefficient $({\mu }_s)$ is used to break the stacity (so when the object is still), while the second one $({\mu }_d)$ is used to determine the friction force that slows down the motion.

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• ~Example:
  • Wood on wood: ${\mu }_s=0.25\sim 0.5$, ${\mu }_d=0.2$
  • Glass on glass: ${\mu }_s=0.94$, ${\mu }_d=0.4$
  • Ski on dry snow: ${\mu }_s\simeq 0$, ${\mu }_d=0.04$
  • When the object is moving and ${\mu }_s\gg {\mu }_s$ (usually true) we approximate the static friction coefficient ${\mu }_s$ to 0, so we do not need to calculate it.
  • Basically when we define the friction cone, the angle is equal to:$\theta ={\tan }^{-1}({\mu }_s)\text{or}\theta ={\tan }^{-1}({\mu }_d)$And it depends if the grasp is static or if the object is sliding in the hand.

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