\section{Theoretical Background} The theoretical background is everything related to the physics part of the project. It covers the calculating the inertia of different types of polygons; different algorithms to detect whether there is a collision between two polygons; the resolution of the collision, i.e. finding the final velocity vectors and angular speed of those polygons. \subsection{Moment of inertia} 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 are applied to it. An object with a lot of inertia requires more force to change its motion, either to make it move if it's at rest or to stop it if it's already moving. On the other hand, an object with less inertia is easier to set in motion or bring to a halt. The moment of inertia is similar but is used in a slightly different context, it specifically refers to the rotational inertia of an object. It measures an object's resistance to changes in its rotational motion and how its mass is distributed with respect to is axis of rotation. In the case of this project the axis of rotation is the one along the $z$-axis (perpendicular to the plane of the simulation) and placed at the barycenter of the polygon. The general formula for the moment of inertia is \begin{equation} \label{eq:moment_general} I_Q = \int \vec r^2 \rho(\vec r) \diff \mathcal{A} \end{equation} where $\rho$ is the density of object $Q$ in the point $\vec r$ across the small pieces of area $\mathcal A$ of the object. In our case, since we are implementing a 2D engine we can use the $\mathbb{R}^2$ coordinate systems, thus the formula becomes $$ I_Q = \iint \rho(x, y) \vec r^2 \diff x\diff y$$ and since the requirements express that the mass of the polygons is spread uniformly across its surface, the formula finally becomes \begin{equation} \label{eq:moment} I_Q = \rho \iint x^2 + y^2 \diff x\diff y \end{equation} The bounds of the integral depend on the shape of the polygon. In the following sections, we will describe how to compute those bounds, then we will show a different technique to compute the moment of inertia of arbitrary polygons. \subsubsection{Rectangle} The moment of inertia of a rectangle of width $w$ and height $h$ with respect to the axis of rotation that passes through its barycenter can be visualized in the \figref{fig:rectangle_inertia}. \begin{figure}[H] \centering \hfill \begin{subfigure}[]{.4\textwidth} \centering \inputtikz{rectangle_inertia2d} \caption{2d view of rectangle with axis of rotation} \label{fig:rectangle_inertia2d} \end{subfigure} \hfill \begin{subfigure}[]{.4\textwidth} \centering \inputtikz{rectangle_inertia3d} \caption{3d view of rectangle with axis of rotation} \label{fig:rectangle_inertia3d} \end{subfigure} \hfill\null \caption{Representation of rectangle with respect to axis of rotation $z$} \label{fig:rectangle_inertia} \end{figure} As figure \figref{fig:rectangle_inertia2d} implies, the bounds of equation \ref{eq:moment} are trivial to derive: \begin{equation} \label{eq:rect_moment} I_{\text{rect}} = \rho \int_{-\frac{h}{2}}^{\frac{h}{2}} \int_{-\frac{w}{2}}^{\frac{w}{2}} x^2 + y^2 \diff x \diff y = \frac{\rho wh}{12} \left( w^2 + h^2 \right) \end{equation} and since $\rho w h$ is the density of the rectangle multiplied by its area, we can replace this term by its mass $m$, thus \begin{equation} I_{\text{rect}} = \frac{1}{12} m\left(w^2 + h^2\right) \end{equation} All the steps to compute equation~\ref{eq:rect_moment} can be found in equation \ref{eq:rect_moment_long} in Appendix \ref{appendix:calculations}. \subsubsection{Regular Polygons} \label{sub:regular_polygons} A regular polygon is a shape that has sides of equal length and angles between those sides of equal measure. A polygon of $n$ sides can be subdivided in $n$ congruent (and isosceles since they are all the radius of the circumscribing circle) triangles that all meet in the polygon's barycenter, as demonstrated in Figure \ref{fig:pentagon_triangles} with a pentagon. \begin{figure}[H] \centering \hfill \begin{subfigure}[]{.45\textwidth} \centering \inputtikz[.6]{pentagon} \caption{Regular polygon of 5 sides with its barycenter} \label{fig:pentagon} \end{subfigure} \hfill \begin{subfigure}[]{.45\textwidth} \centering \inputtikz[.6]{pentagon_congruent} \caption{Pentagon divided in 5 congruent triangles} \label{fig:pentagon_triangles} \end{subfigure} \hfill\null \caption{Subdivision of regular polygons into congruent triangles} \label{fig:regular_poly} \end{figure} If we define one of the sub-triangle of the regular polygon as $T$, then we can find the moment of inertia $I_T$ when it is rotating about the barycenter. To find the bounds of the integral in equation \ref{eq:moment}, we can take the triangle $T$ and place it along the $x$-axis so that it is symmetric likes shown in figure. Assuming the side length of the polygon is $l$, the height of the triangle $T$ is $h$ and the angle of the triangle on the barycenter of the polygon to be $\theta$, then \begin{figure}[H] \centering \inputtikz[.3]{isosceles} \caption{Sub-triangle $T$ of regular polygon} \label{fig:subtriangle} \end{figure} we can see the bounds for the integral \begin{equation} \label{eq:subtriangle_moment} I_{T} = \rho \int_0^h\int_{-\frac{lx}{2h}}^{\frac{lx}{2h}}x^2 + y^2 \diff y\diff x= \frac{m_Tl^2}{24} \left(1 + 3\cot^2\left(\frac{\theta}{2}\right)\right) \end{equation} All the steps to compute equation~\ref{eq:subtriangle_moment} can be found in equation \ref{eq:subtriangle_moment_long} in Appendix \ref{appendix:calculations}. Now that we have the moment of inertia of the sub-triangle, we can make the link to the overall polygon. Since $$ \theta = \frac{2\pi}{n} \implies \frac{\theta}{2} = \frac{\pi}{n} $$ and the moment of inertia are additive (as long they are as they are about the same axis) we can get the moment of inertia with $$ I_{\text{regular}} = n I_T $$ and since the mass of the regular polygon $m$ is the sum of the masses of the sub-triangle $$ m = n m_T $$ we have that \begin{equation} \label{eq:regular_moment} I_{\text{regular}} = \frac{ml^2}{24} \left( 1 + 3\cot^2\left(\frac{\pi}{n}\right) \right) \end{equation} \subsubsection{Arbitrary Polygons} For arbitrary polygons, we are taking a slightly different approach. Using the Cartesian coordinate system to solve the equation \ref{eq:moment} revealed to be more cumbersome than useful. But similarly to regular polygons (c.f. Section \ref{sub:regular_polygons}), we can use the additive property of the moment inertia to divide our arbitrary polygon into sub-triangles. As opposed to regular polygons, these triangles won't be congruent, so we can't just get the moment of inertia of one of them and multiply it by the number of sides, but we need to calculate them individually. So given a polygon of $n$ sides, we can construct $n$ sub-triangles $T_i$, for $i = 1, \dots, n$. So the moment of inertia $I$ of the polygon will be \begin{equation} I = \sum_i I_{T_i} \end{equation} \begin{figure}[H] \centering \begin{subfigure}[]{.5\textwidth} \centering \inputtikz[.7]{arbitrary} \caption{An arbitrary 6-sided polygon} \label{fig:arbitrary} \end{subfigure} \begin{subfigure}[]{.49\textwidth} \centering \inputtikz[.7]{arbitrary_divided} \caption{Arbitrary polygon divided into 6 sub-triangles} \label{fig:abitrary_divded} \end{subfigure} \end{figure} To calculate the moment of inertia $I_{T_i}$, instead of using the classical $x$- and $y$-axis as we did before, we decided to use the edges of the triangle as axis and therefore express what we need to integrate in function of those as can be seen in Figure \ref{fig:abitrary_subtriangle}. \begin{figure}[H] \centering \inputtikz[.7]{arbitrary_subtriangle} \caption{Sub-triangle of arbitrary polygon} \label{fig:abitrary_subtriangle} \end{figure} In Figure \ref{fig:abitrary_subtriangle}, $C$ represent the barycenter of the polygon (as is shown in Figure \ref{fig:abitrary_divded}). The axis we are going to integrate on are $\vv{CA}$ and $\vv{AB}$. We can now define \begin{equation} \label{eq:alpha} \vv{CP_1} = \alpha \vv{CA}, \qquad \vv{CP_2} = \alpha \vv{CB}, \qquad \forall \alpha \in [0, 1] \end{equation} and $$ \vv{P_1Q} = \beta \vv{P_1P_2}, \qquad \forall \beta \in [0, 1] $$ From \ref{eq:alpha}, it quickly follows that \[ \vv{P_1P_2} = \alpha \vv{AB} \] therefore \begin{equation} \label{eq:beta_alpha} \vv{P_1Q} = \beta \alpha \vv{AB} \end{equation} Finally, if we put together equations \ref{eq:alpha} and \ref{eq:beta_alpha}, we have that \begin{equation}\label{eq:r} \vec r = \vv{CP_1} + \vv{P_1Q} = \alpha \vv{CA} + \beta \alpha \vv{AB} \end{equation} Now we got the first part equation \ref{eq:moment_general}. To find the $\diff \mathcal A$, we just need to get the area of the square that contains $Q$ in Figure \ref{fig:abitrary_subtriangle}. Since $\|\vv{AB}\|$ represents the base of the triangle $T_i$, we can define $$ b = \| \vv{AB}\| $$ we consequently have that \begin{equation} \label{eq:dA} \diff \mathcal{A} = b \alpha \diff \beta h\diff \alpha \end{equation} where $h = \| \vv{CH} \|$ is the height of triangle. We can now assemble \ref{eq:r} and \ref{eq:dA} \begin{equation} \label{eq:subtriangle_arbitrary_moment} I_{T_i} = \rho \int_0^1 \int_0^1 \vec r^2 hb \alpha \diff \alpha \diff \beta = \frac{\rho h b}{4} \left(\frac{1}{3} \vv{AB}^2 + \vv{AB} \cdot \vv{CA} + \vv{CA}^2\right) \end{equation} Since $\frac{\rho h b}{2}$ is the mass of the triangle we can write the result as \begin{equation} I_{T_i} = \frac{m_{T_i}}{2} \left(\frac{1}{3} \vv{AB}^2 + \vv{AB} \cdot \vv{CA} + \vv{CA}^2\right) \end{equation} All the steps to compute equation~\ref{eq:subtriangle_arbitrary_moment} can be found in equation \ref{eq:subtriangle_arbitrary_moment_long} in Appendix \ref{appendix:calculations}. Now that we have the moment of inertia of the sub-triangle, we can make the link to the overall polygon. \begin{equation} I_{\text{arbitrary}} = \sum_i I_{T_i} = \sum_{i=1}^n \frac{m_{T_i}}{2} \left(\frac{1}{3} \vv{P_iP_{i+1}}^2 + \vv{CP_i} \cdot \vv{P_iP_{i+1}} + \vv{CP_i}^2\right) \end{equation} where, $P_{n+1} = P_1$ in the case of $i = n$. \subsection{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 impact point and a normal vector, that will then be used for the collision resolution \ref{sub:resolution}. \subsubsection{Separating Axis Theorem} This algorithm was the first one studied for this project and was inspired by the works of David Eberly \cite{convexcollisionsSAT}. The separating axis theorem (SAT) states that if you can draw a line between two convex objects, they do not overlap. We will call this line a \textit{separating line}. More technically, two convex shapes do not overlap if there exists an axis onto which the two objects' projections do not overlap. We'll call this axis a \textit{separating axis}. This concept can be visualized in Figure \ref{fig:SAT-intro}. \begin{figure}[H] \centering \inputtikz[.7]{SAT_intro} \caption{SAT: Separating axis ($A$) vs non-separating axis ($B$), with separating line ($C$)} \label{fig:SAT-intro} \end{figure} As we can see in Figure \ref{fig:SAT-intro}, the axis $B$ show that the projections of the both polygons overlap, but we were able to find an axis $A$ where this is not the case. As soon as we find an axis for which the projections do not overlap, it means that the polygons are not colliding. For 2D objects, we only need to consider the axes that are orthogonal to each edge. In Figure \ref{fig:SAT-intro}, only two of those axes are shown for better readability, but they would be 7, one for each edge. To move (or push) one polygon away from the other, we also need to find a vector that, when added to the polygons position, will make the shapes not overlap. We want the minimum displacement possible, We'll call this vector the minimum push vector (MPV). For 2-dimensional polygons, this vector will lie in some of the orthogonal axes. \begin{figure}[H] \centering \inputtikz[.7]{SAT_mvp} \caption{SAT: Minimum push vector $\vec v_i$ on axis defined by $\vec o_i$, orthogonal to edge $e_i$} \label{fig:SAT-mpv} \end{figure} The candidate MPVs $\vec v_i$ are the vectors that define the axis $\vec o_i$ (orthogonal to edge $e_i$), with $\| \vec o_i \| = 1$, multiplied by the minimum overlap between the two polygons, as shown in \ref{fig:SAT-mpv}. The final MPV is simply the $\vec v_i$ with the smallest norm. \paragraph{Pitfalls of SAT} The issue with the SAT algorithm is that although it is good to find whether two polygons are colliding and the MPV, it isn't trivial to gather the point of impact, i.e. the vertex that is penetrating the other polygon. It is doable, but during the implementation, it came with some caveats that introduced some bugs, so we decided to switch strategy and go with an algorithm of our own. Moreover, SAT only supports convex polygons, which limits the original objective of the project, which was to have any arbitrary polygon. \subsubsection{Vertex collisions} The solution that was adopted for the project, after trying SAT, was a more intuitive one, developed by Prof. Carzaniga. The idea is simple: check if a vertex of a polygon is colliding with an edge of another polygon. \begin{figure}[H] \centering \inputtikz[.5]{vertex_intro} \caption{Vertex-edge collision between polygons $A$ and $B$} \label{fig:vertex-edge} \end{figure} If we have a polygon defined as a set of points $P \subset \mathbb{R}^2$, we define a vertex as a pair of segments $\left(\overline{V_{i-1}V_i}, \overline{V_{i+1}V_i}\right)$. To check if vertex $V_i$ of polygon $A$ is inside polygon $B$, we just check if the both segments $\overline{V_{i-1}V_i}$ and $\overline{V_{i+1}V_i}$ intersect edge $e_j$. If such is the case, we know that $V_i$ is inside, and we can use it as impact point. We can now take the vector perpendicular to the edge $e_j$ and normalize it, which is the normal we need for the collision resolution. \paragraph{Edge cases} With this approach (and many other collision detection algorithms), it is easy to see that the case described in the Figure \ref{fig:vertex-edge} is only a general one. It will ultimately happen most often, but there are especially two edge cases that are note-worthy and occurred during the simulation a greater number of times than expected. \paragraph{Parallel collision} Parallel collision occur when two edge collide with each other, an example can be seen in Figure \ref{fig:vertex-parallel}. This collision does not get detected by the vertex-edge detection method because the edge of $A$ parallel to the edge of $B$ do not collide, since they are parallel. \begin{figure}[H] \centering \inputtikz[.5]{vertex_parallel} \caption{Parallel edges collision between polygons $A$ and $B$} \label{fig:vertex-parallel} \end{figure} To determine if polygon $A$ is having a parallel collision with polygon $B$ we check if only the segment $\overline{V_{i-1}V_i}$ intersects with edge $e_j$ and if the segment $\overline{V_{i+1}V_i}$ is parallel to edge $e_j$. To find the impact point, we first need to find the minimum overlap between the parallel edges. We can calculate it by projecting the points that make the parallel edges of $A$ and $B$ on the axis generated by the edge of $A$ (c.f. Figure \ref{fig:parallel-impact}).Finally, the impact point, is the midpoint of the said overlap. \begin{figure}[H] \centering \begin{subfigure}[]{.5\textwidth} \centering \inputtikz[.7]{vertex_parallel_full} \caption{Edge fully contained by other edge} \end{subfigure} \begin{subfigure}[]{.49\textwidth} \centering \inputtikz[.7]{vertex_parallel_partial} \caption{Edge partly contained by other edge} \end{subfigure} \caption{Parallel collision, finding the impact point} \label{fig:parallel-impact} \end{figure} The normal vector is given in the same way as the vertex-edge collision, i.e. taking the perpendicular vector to~$e_j$ and normalizing it. % TODO: ask Carza about the possibility of resolving vertex-vertex normal with SAT \paragraph{Vertex on vertex collision} These collision happen when, at the moment of the frame, two polygons are "inside each other", i.e. they both have have a vertex present inside the area of the other polygon, as shown in Figure \ref{fig:vertex-vertex}. \begin{figure}[H] \centering \inputtikz[.5]{vertex_vertex} \caption{Vertex on vertex collision between polygons $A$ and $B$} \label{fig:vertex-vertex} \end{figure} \noindent Two problems arise when trying to deal with this edge case: \begin{enumerate} \item determining which of the two vertices we chose as the impact point; \label{itm:impact} \item determining what normal vector. \label{itm:normal} \end{enumerate} % \noindent For point \ref{itm:impact}, it actually doesn't really matter. What is represented in Figures~\ref{fig:vertex-edge}-\ref{fig:vertex-vertex} are an over exaggeration of what happens in the engine. Since the time delta between two frames is so small, the collisions look more like what is in Figure~\ref{fig:real-vertex-vertex}, where the concerned vertices are located practically in the same spot, and the difference is negligible. So for convenience with the general vertex-edge case, we will take the vertex of $A$ being inside $B$ as the impact point. For point \ref{itm:normal}, that's where the real problem of the edge case occur. We are unable to determine what vectors is supposed to be the normal vector. One might say "just take the vector that goes from the vertex of $A$ (that is inside $B$) to the one of $B$ (that is inside $A$)", but it doesn't work out in general. It works (kind of) when we have non-rotating objects going straight at each other, but as soon as you have rotational motion, this reasoning collapses. The solution that was decided (together with advisor Prof. Carzaniga) was to treat the collision as a vertex-edge collision, and choosing whatever the first edge of $B$ comes up first in the calculations as the edge to find the normal. The results look realistic enough to be accepted. \begin{figure}[H] \centering \begin{subfigure}[]{.33\textwidth} \centering \inputtikz[.7]{vertex_edge_real} \caption{Realistic vertex-edge collision} \end{subfigure} \begin{subfigure}[]{.32\textwidth} \centering \inputtikz[.7]{vertex_parallel_real} \caption{Realistic edge-edge collision} \end{subfigure} \begin{subfigure}[]{.34\textwidth} \centering \inputtikz[.7]{vertex_vertex_real} \caption{Realistic vertex-vertex collision} \label{fig:real-vertex-vertex} \end{subfigure} \caption{Realistic collisions} \label{fig:real-collisions} \end{figure} \subsection{Collision resolution} \label{sub:resolution}