Written the collision resolution section

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Karma Riuk 2023-06-14 11:59:21 +02:00
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\usepackage[]{subcaption}
\usepackage[]{float}
\usepackage[]{multicol}
\usepackage{mathtools}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetikzlibrary{external,shapes,through}
\usetikzlibrary{external,shapes,through,arrows}
\tikzexternalize[prefix=figures/]
\tikzstyle{none}=[]
\input{./tikzs/styles.tikzstyles}
\tikzset{>=stealth}
\pgfdeclarelayer{nodelayer}
\pgfdeclarelayer{edgelayer}
\pgfsetlayers{edgelayer,nodelayer,main}
\graphicspath{{../figures/}{./figures/}}
\newcommand*{\vv}[1]{\overrightarrow{#1}}
\newcommand*{\figref}[1]{\figurename~\ref{#1}}
\captionsetup{labelfont={bf}}

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@ -3,7 +3,7 @@
\section{Calculations}
\label{appendix:calculations}
\paragraph{Moment of inertia of rectangle}
\subsection{Moment of inertia of rectangle}
\begin{equation}
\label{eq:rect_moment_long}
\begin{split}
@ -22,7 +22,7 @@
\end{equation}
\newpage
\paragraph{Moment of inertia of sub-triangle of regular polygon}
\subsection{Moment of inertia of sub-triangle of regular polygon}
Before starting the calculations, it is to be noted that according to Figure
\ref{fig:subtriangle}, we have that
$$ \tan\left(\frac{\theta}{2}\right) = \frac{\frac{l}{2}}{h} = \frac{l}{2h} $$
@ -46,7 +46,7 @@ it will be useful to simplify the result of the integral.
\end{equation}
\newpage
\paragraph{Moment of inertia of sub-triangle of arbitrary polygon} Recall
\subsection{Moment of inertia of sub-triangle of arbitrary polygon} Recall
equation \ref{eq:r} defines
$$ \vec r = \alpha \vv{CA} + \beta \alpha \vv{AB} $$
\begin{equation}
@ -62,3 +62,10 @@ $$ \vec r = \alpha \vv{CA} + \beta \alpha \vv{AB} $$
&= \frac{\rho hb}{4} \left(\frac{1}{3}\vv{AB}^2 + \vv{AB} \cdot \vv{CA} + \vv{CA}^2\right) \\
\end{split}
\end{equation}
\newpage
\subsection{Solving for impulse parameter}
\label{app:impulse_long}
[to be done :)]
\begin{equation}
\end{equation}

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@ -6,6 +6,7 @@ polygons; the resolution of the collision, i.e. finding the final
velocity vectors and angular speed of those polygons.
\subsection{Moment of inertia}
\label{sub:moment}
The inertia of an object refers to the tendency of an object to resist a change
of its state of motion or rest, it describes how the object behaves when forces
@ -264,6 +265,7 @@ to the overall polygon.
where, $P_{n+1} = P_1$ in the case of $i = n$.
\subsection{Collision detection}
\label{sub:collision-detection}
Collision detection, as the name suggests, are the algorithms used to detect
whether two polygons are colliding. The result of this procedure must be an
@ -462,3 +464,161 @@ find the normal. The results look realistic enough to be accepted.
\subsection{Collision resolution}
\label{sub:resolution}
The collision resolution is the last step in the processing of the collision.
Algorithmically, it is much less heavy than collision detection, since, once the
simulation has two colliding polygons, a point of impact and a normal vector,
it's just a case of applying the rigid body physics formulas to the polygons
that are colliding. This part has been helped a lot by the works of Erik Neumann
\cite{collision:resolution-site} and Chris Hecker
\cite{collision:resolution-paper}.
\begin{figure}[H]
\centering
\inputtikz[.5]{collision_resolution}
\caption{Collision resolution between polygons $A$ and $B$}
\label{fig:collision_resolution}
\end{figure}
\paragraph{Variable definition} Before getting into any maths, let's define some
variables that we are going to use
\begin{multicols}{2}
\begin{itemize}
\item $m_a, m_b = $ mass of the bodies $A$ and $B$
\item $\vec r_{ap} = $ distance vector from center of mass of body $A$ to
point $P$
\item $\vec r_{bp} = $ distance vector from center of mass of body $A$ to
point $P$
\item $\omega_{a1}, \omega_{b1} = $ initial angular velocity of bodies
$A, B$
\item $\omega_{a2}, \omega_{b2} = $ final angular velocity of bodies
$A, B$
\item $\vec v_{a1}, \vec v_{b1} =$ initial velocities of center of mass
bodies $A, B$
\item $\vec v_{a2}, \vec v_{b2} =$ final velocities of center of mass
bodies $A, B$
\item $\vec v_{ap1}=$ initial velocity of impact point $P$ on body $A$
\item $\vec v_{bp1}=$ initial velocity of impact point $P$ on body $B$
\item $\vec v_{p1}=$ initial relative velocity of impact points on body $A, B$
\item $\vec v_{p2}=$ final relative velocity of impact points on body $A, B$
\item $\vec n=$ normal vector
\item $ e=$ elastic coefficient (0 = inelastic, 1 = perfectly elastic)
\end{itemize}
\end{multicols}
Some of those variables, like the ones in the left column, are already given by
the simulation, some of them have been computed thanks to the algorithms in
section \ref{sub:collision-detection} (normal vector and position of point $P$),
and some have still to be defined mathematically, such as $\vec v_{ap1},\vec
v_{bp1},\vec v_{p1}$ and $\vec v_{p2}$. So the velocity vectors of impact
point $P$ on both bodies, before the collision, are
\begin{equation} \label{eq:vabp1}
\begin{split}
\vec v_{ap1} = \vec v_{a1} + \omega_{a1} \times \vec r_{ap} \\
\vec v_{bp1} = \vec v_{b1} + \omega_{b1} \times \vec r_{bp}
\end{split}
\end{equation}
Similarly, the final velocities are
\begin{equation} \label{eq:vabp2}
\begin{split}
\vec v_{ap2} = \vec v_{a2} + \omega_{a2} \times \vec r_{ap}\\
\vec v_{bp2} = \vec v_{b2} + \omega_{b2} \times \vec r_{bp}
\end{split}
\end{equation}
Here we are regarding the angular velocity as a 3-dimensional vector
perpendicular to the plane, so that the cross product is calculated as
$$ \omega \times \vec r = \begin{pmatrix} 0\\0\\\omega \end{pmatrix} \times
\begin{pmatrix} r_x\\r_y\\0 \end{pmatrix} = \begin{pmatrix} -\omega r_y \\
\omega r_x \\0\end{pmatrix} $$
We these variables, we can finally define the relative velocities $\vec
v_{p1}$ and $\vec v_{p2}$
\[ \begin{split}
\vec v_{p1} = \vec v_{ap1} - \vec v_{bp2}\\
\vec v_{p2} = \vec v_{ap2} - \vec v_{bp2}
\end{split} \]
If we expand by using \ref{eq:vabp1} and \ref{eq:vabp2}, we get
\begin{equation}
\begin{split}
\vec v_{p1} = \vec v_{a1} + \omega_{a1} \times \vec r_{ap} - \vec v_{b1}
+ \omega_{b1} \times \vec r_{bp}\\
\vec v_{p2} = \vec v_{a2} + \omega_{a2} \times \vec r_{ap} - \vec v_{b2} + \omega_{b2} \times \vec r_{bp}
\end{split}
\end{equation}
The relative velocity of point $P$ along the normal vector $\vec n$ is
$$ \vec v_{p1} \cdot \vec n $$
Note that for a collision to occur this relative normal velocity must be
negative (that is, the objects must be approaching each other). Let e be the
elasticity of the collision, having a value between 0 (inelastic) and 1
(perfectly elastic). We now make an important assumption in the form of the
following relation
\begin{equation}
\label{eq:vp2n}
\vec v_{p2} \cdot \vec n = - e \vec v_{p1} \cdot \vec n
\end{equation}
This says that the velocity at which the objects fly apart is proportional to
the velocity with which they were coming together. The proportionality factor is
the elasticity $e$.
\paragraph{Collision Impulse} In simple terms, collision impulse refers to the
change in momentum experienced by an object during a collision. It is a measure
of the force applied to an object over a short period of time. We imagine that
during the collision there is a very large force acting for a very brief period
of time. If you integrate (sum) that force over that brief time, you get the
impulse.
In our simulation, we are assuming that no friction is happening during the
collision, so that the only force we have to consider is the one of along the
normal vector $\vec n$. The friction would create a force perpendicular to the
normal, and it would make things a little too complicated for the scope of this
project.
Since the only force we consider is the normal one, we can consider the net
impulse to be $j \vec n$, where $j$ is the impulse parameter. The body $A$ will
experience the net impulse $j \vec n$ and body $B$ will experience it's negation
$- j \vec n$ since the force that $B$ experience is equal an opposite to the one
experienced by $A$. The impulse is a change in momentum. Momentum has units of
velocity times mass, so if we divide the impulse by the mass we get the change
in velocity. We can relate initial and final velocities as
\begin{equation}
\label{eq:va2}
\vec v_{a2} = \vec v_{a1} + \frac{j\vec n}{m_a}
\end{equation}
\begin{equation}
\label{eq:vb2}
\vec v_{b2} = \vec v_{b1} - \frac{j\vec n}{m_b}
\end{equation}
For the final angular speed, it's change from the impulse $j\vec n$ is given by
$\vec r_{ap} \times j \vec n$. We can divide the result by the moment of
inertia, which was calculated in section \ref{sub:moment}, to get convert the
change in angular momentum into change in angular speed. Similarly as above, we
can relate the initial and final angular velocity as
\begin{equation}
\label{eq:omega_a2}
\omega_{a2} = \omega_{a1} + \frac{\vec r_{ap} \times j\vec n}{I_a}
\end{equation}
\begin{equation}
\label{eq:omega_b2}
\omega_{b2} = \omega_{b1} - \frac{\vec r_{bp} \times j\vec n}{I_b}
\end{equation}
\paragraph{Solving for the impulse parameter}
Now we have everything we need to solve for the impulse parameter $j$. The bulk
of the calculations can be found in section \ref{app:impulse_long} of the
appendix \ref{appendix:calculations}. But basically we start with equation
\ref{eq:vp2n} and we end up with
\begin{equation}
\label{eq:j}
j = \frac{ - (1+e) \cdot \vec v_{ap1} \cdot \vec n }{\frac{1}{m_a} + \frac{1}{m_b} +
\frac{\left( \vec r_{ap} \times \vec n \right)^2}{I_a} + \frac{\left( \vec
r_{bp} \times \vec n \right)^2}{I_b}}
\end{equation}