275 lines
10 KiB
TeX
275 lines
10 KiB
TeX
\section{Theoretical Background}
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The theoretical background is everything related to the physics part of the
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project. It covers the calculating the inertia of different types of polygons;
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different algorithms to detect whether there is a collision between two
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polygons; the resolution of the collision, i.e. finding the final
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velocity vectors and angular speed of those polygons.
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\subsection{Moment of inertia}
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The inertia of an object refers to the tendency of an object to resist a change
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of its state of motion or rest, it describes how the object behaves when forces
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are applied to it. An object with a lot of inertia requires more force to change
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its motion, either to make it move if it's at rest or to stop it if it's already
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moving. On the other hand, an object with less inertia is easier to set in
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motion or bring to a halt.
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The moment of inertia is similar but is used in a slightly different context, it
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specifically refers to the rotational inertia of an object. It measures an
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object's resistance to changes in its rotational motion and how its mass is
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distributed with respect to is axis of rotation.
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In the case of this project the axis of rotation is the one along the $z$-axis
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(perpendicular to the plane of the simulation) and placed at the barycenter of
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the polygon.
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The general formula for the moment of inertia is
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\begin{equation}
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\label{eq:moment_general}
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I_Q = \int \vec r^2 \rho(\vec r) \diff \mathcal{A}
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\end{equation}
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where $\rho$ is the density of object $Q$ in the point $\vec r$ across the
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small pieces of area $\mathcal A$ of the object.
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In our case, since we are implementing a 2D engine we can use the $\mathbb{R}^2$
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coordinate systems, thus the formula becomes
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$$ I_Q = \iint \rho(x, y) \vec r^2 \diff x\diff y$$
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and since the requirements express that the mass of the polygons is spread
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uniformly across its surface, the formula finally becomes
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\begin{equation}
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\label{eq:moment}
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I_Q = \rho \iint x^2 + y^2 \diff x\diff y
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\end{equation}
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The bounds of the integral depend on the shape of the polygon. In the following
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sections, we will describe how to compute those bounds, then we will show a
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different technique to compute the moment of inertia of arbitrary polygons.
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\subsubsection{Rectangle}
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The moment of inertia of a rectangle of width $w$ and height $h$ with respect to
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the axis of rotation that passes through its barycenter can be visualized in the
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\figref{fig:rectangle_inertia}.
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\begin{figure}[H]
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\centering
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\hfill
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\begin{subfigure}[]{.4\textwidth}
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\centering
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\inputtikz{rectangle_inertia2d}
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\caption{2d view of rectangle with axis of rotation}
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\label{fig:rectangle_inertia2d}
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\end{subfigure}
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\hfill
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\begin{subfigure}[]{.4\textwidth}
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\centering
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\inputtikz{rectangle_inertia3d}
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\caption{3d view of rectangle with axis of rotation}
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\label{fig:rectangle_inertia3d}
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\end{subfigure}
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\hfill\null
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\caption{Representation of rectangle with respect to axis of rotation $z$}
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\label{fig:rectangle_inertia}
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\end{figure}
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As figure \figref{fig:rectangle_inertia2d} implies, the bounds of equation
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\ref{eq:moment} are trivial to derive:
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\begin{equation}
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\label{eq:rect_moment}
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I_{\text{rect}} = \rho \int_{-\frac{h}{2}}^{\frac{h}{2}}
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\int_{-\frac{w}{2}}^{\frac{w}{2}} x^2 + y^2 \diff x \diff y
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= \frac{\rho wh}{12} \left( w^2 + h^2 \right)
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\end{equation}
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and since $\rho w h$ is the density of the rectangle multiplied by its area, we
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can replace this term by its mass $m$, thus
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\begin{equation}
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I_{\text{rect}} = \frac{1}{12} m\left(w^2 + h^2\right)
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\end{equation}
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All the steps to compute equation~\ref{eq:rect_moment} can be found in equation
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\ref{eq:rect_moment_long} in Appendix \ref{appendix:calculations}.
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\subsubsection{Regular Polygons}
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\label{sub:regular_polygons}
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A regular polygon is a shape that has sides of equal length and angles between
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those sides of equal measure. A polygon of $n$ sides can be subdivided in $n$
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congruent (and isosceles since they are all the radius of the circumscribing
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circle) triangles that all meet in the polygon's barycenter,
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as demonstrated in Figure \ref{fig:pentagon_triangles} with a pentagon.
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\begin{figure}[H]
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\centering
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\hfill
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\begin{subfigure}[]{.45\textwidth}
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\centering
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\inputtikz[.6]{pentagon}
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\caption{Regular polygon of 5 sides with its barycenter}
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\label{fig:pentagon}
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\end{subfigure}
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\hfill
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\begin{subfigure}[]{.45\textwidth}
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\centering
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\inputtikz[.6]{pentagon_congruent}
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\caption{Pentagon divided in 5 congruent triangles}
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\label{fig:pentagon_triangles}
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\end{subfigure}
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\hfill\null
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\caption{Subdivision of regular polygons into congruent triangles}
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\label{fig:regular_poly}
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\end{figure}
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If we define one of the sub-triangle of the regular polygon as $T$, then we can
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find the moment of inertia $I_T$ when it is rotating about the barycenter. To
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find the bounds of the integral in equation \ref{eq:moment}, we can take the
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triangle $T$ and place it along the $x$-axis so that it is symmetric likes shown
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in figure. Assuming the side length of the polygon is $l$, the height of the
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triangle $T$ is $h$ and the angle of the triangle on the barycenter of the
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polygon to be $\theta$, then
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\begin{figure}[H]
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\centering
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\inputtikz[.3]{isosceles}
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\caption{Sub-triangle $T$ of regular polygon}
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\label{fig:subtriangle}
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\end{figure}
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we can see the bounds for the integral
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\begin{equation}
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\label{eq:subtriangle_moment}
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I_{T} = \rho \int_0^h\int_{-\frac{lx}{2h}}^{\frac{lx}{2h}}x^2 + y^2 \diff
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y\diff x= \frac{m_Tl^2}{24} \left(1 + 3\cot^2\left(\frac{\theta}{2}\right)\right)
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\end{equation}
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All the steps to compute equation~\ref{eq:subtriangle_moment} can be found in
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equation \ref{eq:subtriangle_moment_long} in Appendix
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\ref{appendix:calculations}.
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Now that we have the moment of inertia of the sub-triangle, we can make the link
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to the overall polygon. Since
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$$ \theta = \frac{2\pi}{n} \implies \frac{\theta}{2} = \frac{\pi}{n} $$
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and the moment of inertia are additive (as long they are as they are about the
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same axis) we can get the moment of inertia with
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$$ I_{\text{regular}} = n I_T $$
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and since the mass of the regular polygon $m$ is the sum of the masses of the
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sub-triangle
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$$ m = n m_T $$
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we have that
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\begin{equation}
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\label{eq:regular_moment}
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I_{\text{regular}} = \frac{ml^2}{24} \left( 1 + 3\cot^2\left(\frac{\pi}{n}\right) \right)
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\end{equation}
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\subsubsection{Arbitrary Polygons}
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For arbitrary polygons, we are taking a slightly different approach. Using the
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Cartesian coordinate system to solve the equation \ref{eq:moment} revealed to be
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more cumbersome than useful. But similarly to regular polygons (c.f. Section
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\ref{sub:regular_polygons}), we can use the additive property of the moment
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inertia to divide our arbitrary polygon into sub-triangles. As opposed to
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regular polygons, these triangles won't be congruent, so we can't just get the
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moment of inertia of one of them and multiply it by the number of sides, but we
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need to calculate them individually. So given a polygon of $n$ sides, we can
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construct $n$ sub-triangles $T_i$, for $i = 1, \dots, n$. So the moment of
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inertia $I$ of the polygon will be
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\begin{equation}
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I = \sum_i I_{T_i}
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\end{equation}
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\begin{figure}[H]
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\centering
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\begin{subfigure}[]{.5\textwidth}
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\centering
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\inputtikz[.7]{arbitrary}
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\caption{An arbitrary 6-sided polygon}
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\label{fig:arbitrary}
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\end{subfigure}
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\begin{subfigure}[]{.49\textwidth}
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\centering
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\inputtikz[.7]{arbitrary_divided}
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\caption{Arbitrary polygon divided into 6 sub-triangles}
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\label{fig:abitrary_divded}
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\end{subfigure}
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\end{figure}
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To calculate the moment of inertia $I_{T_i}$, instead of using the classical
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$x$- and $y$-axis as we did before, we decided to use the edges of the triangle
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as axis and therefore express what we need to integrate in function of those as
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can be seen in Figure \ref{fig:abitrary_subtriangle}.
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\begin{figure}[H]
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\centering
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\inputtikz[.7]{arbitrary_subtriangle}
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\caption{Sub-triangle of arbitrary polygon}
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\label{fig:abitrary_subtriangle}
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\end{figure}
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In Figure \ref{fig:abitrary_subtriangle}, $C$ represent the barycenter of the
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polygon (as is shown in Figure \ref{fig:abitrary_divded}). The axis we are going
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to integrate on are $\vv{CA}$ and $\vv{AB}$.
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We can now define
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\begin{equation} \label{eq:alpha}
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\vv{CP_1} = \alpha \vv{CA}, \qquad \vv{CP_2} = \alpha
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\vv{CB}, \qquad \forall \alpha \in [0, 1]
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\end{equation}
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and
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$$ \vv{P_1Q} = \beta \vv{P_1P_2}, \qquad \forall \beta \in [0, 1] $$
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From \ref{eq:alpha}, it quickly follows that
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\[
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\vv{P_1P_2} = \alpha \vv{AB}
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\]
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therefore
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\begin{equation} \label{eq:beta_alpha}
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\vv{P_1Q} = \beta \alpha \vv{AB}
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\end{equation}
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Finally, if we put together equations \ref{eq:alpha} and \ref{eq:beta_alpha}, we
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have that
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\begin{equation}\label{eq:r}
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\vec r = \vv{CP_1} + \vv{P_1Q} = \alpha \vv{CA} + \beta \alpha \vv{AB}
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\end{equation}
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Now we got the first part equation \ref{eq:moment_general}. To find the $\diff
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\mathcal A$, we
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just need to get the area of the square that contains $Q$ in Figure
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\ref{fig:abitrary_subtriangle}. Since $\|\vv{AB}\|$ represents the base of the
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triangle $T_i$, we can define
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$$ b = \| \vv{AB}\| $$
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we consequently have that
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\begin{equation} \label{eq:dA}
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\diff \mathcal{A} = b \alpha \diff \beta h\diff \alpha
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\end{equation}
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where $h = \| \vv{CH} \|$ is the height of triangle.
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We can now assemble \ref{eq:r} and \ref{eq:dA}
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\begin{equation}
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\label{eq:subtriangle_arbitrary_moment}
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I_{T_i} = \rho \int_0^1 \int_0^1 \vec r^2 hb \alpha \diff \alpha
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\diff \beta = \frac{\rho h b}{4} \left(\frac{1}{3} \vv{AB}^2 +
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\vv{AB} \cdot \vv{CA} + \vv{CA}^2\right)
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\end{equation}
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Since $\frac{\rho h b}{2}$ is the mass of the triangle we can write the result
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as
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\begin{equation}
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I_{T_i} = \frac{m_{T_i}}{2} \left(\frac{1}{3} \vv{AB}^2 + \vv{AB} \cdot \vv{CA}
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+ \vv{CA}^2\right)
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\end{equation}
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All the steps to compute equation~\ref{eq:subtriangle_arbitrary_moment} can be
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found in equation \ref{eq:subtriangle_arbitrary_moment_long} in Appendix
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\ref{appendix:calculations}.
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Now that we have the moment of inertia of the sub-triangle, we can make the link
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to the overall polygon.
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\begin{equation}
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I_{\text{arbitrary}} = \sum_i I_{T_i} = \sum_{i=1}^n \frac{m_{T_i}}{2}
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\left(\frac{1}{3} \vv{P_iP_{i+1}}^2 + \vv{CP_i} \cdot \vv{P_iP_{i+1}} +
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\vv{CP_i}^2\right)
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\end{equation}
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where, $P_{n+1} = P_1$ in the case of $i = n$.
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\subsection{Collision detection}
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\subsubsection{Separating Axis Theorem}
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\subsubsection{Vertex collisions}
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\subsection{Collision resolution}
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\label{sub:resolution}
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\cite{collision:resolution}
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\subsubsection{Physics}
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\subsubsection{Solving for the impulse parameter}
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