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<h1><a title="physicsforgames.com" href="/">Physics for Games</a></h1>
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<ul class="outer-tree">
<li><a href="/">Introduction</a></li>
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<span>🏀<span>2D</span></span>
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<li><label>Rigidbody</label></li>
<li><a title="/2d/rigidbody/rigidbody_1.html" href="/2d/rigidbody/rigidbody_1.html">Linear Forces</a></li>
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<li><label>Collisions</label></li>
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<li><a title="/2d/_collisions/polygon_polygon.html" href="/2d/_collisions/polygon_polygon.html">Separating Axis Theorem</a></li>
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<span>🌠<span>3D</span></span>
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<li><label>Rigidbody</label></li>
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<span>🔧<span>WebAssembly</span></span>
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<article>
<h1>Separating Axis Theorem</h1>
<section>
<p>
The Separating Axis Theorem (SAT) provides a way to find the intersection between any <i>n</i>-sided <a href='https://ianqvist.blogspot.com/2009/09/convex-polygon-based-collision.html'>convex</a> polygon or circle. In this tutorial, I will explain how this theorem works, and how you can use it to both detect and resolve collisions in your simulation.
</p>
</section>
<section>
<h2>Explanation of Separating Axis Theorem</h2>
<p>
SAT makes use of vector projection to figure out whether or not two concave polygons are intersecting. I believe the best way to describe it is to show you the instance where two polygons do <b>not</b> intersect.
<br/>
<br/>
Images two shapes <b>A</b> and <b>B</b> as shown below (they are triangles here, but they can be anything really).
<br/>
<br />
<div class='image'>
<img src="polygon_polygon/images/1.png" />
</div>
<br/>
We know inutitively that if we were able to draw a line between the two objects, that means that they're not intersecting. In this case, that line is a vertical-ish line:
<br/>
<br />
<div class='image'>
<img src='polygon_polygon/images/2.png' />
</div>
<br/>
That line could be any of the infinite number of lines between the objects, so we can't really select that <i>exact</i> line. So the problem becomes: how do we <i>infer</i> that the vertical line shown in the previous diagram is a line that splits the objects? That is where the Separating Axis Theorem comes in.
<br />
<br/>
To do separating axis theorem, we iterate all of the edges of <b>both</b> shapes. For each edge, we get the normal at that edge (In two-dimensions, this amounts to getting the perpendicular of the line segment defined by the two points of the edge).
<br />
<br />
<div class='image'>
<img src='polygon_polygon/images/2a.png' />
</div>
<br />
We then, in our heads, extend this normal to create an axis across our scene (keep in mind that you won't have to do any "extending" of this axis in code; the normal will do just fine).
<div class='image'>
<img src='polygon_polygon/images/2d.png' />
</div>
<br />
At this point, you will see that I named the three vertices on each of the triangles. Triangle <b>A</b> is made of vertices <i>a1, a2, a3</i> and triangle <b>B</b> is made of vertices <i>b1, b2, b3</i>. At this point, we project the shape onto the axis. To do this, we project the vertices of each shape onto the axis to figure out which one <i>maximizes</i> the dot product and which one <i>minimizes</i> the dot product. The projection of the shape is then the range defined by this min and max value.
<br />
<br />
<div class='image'>
<img src='polygon_polygon/images/2e.png' />
</div>
<br />
Finally, we check if the two ranges overlap. If they don't overlap, then the shapes do <b>not</b> intersect. If they do overlap, we repeat the process for all of the other edges. If all of these ranges overlap, then the shape overlaps, and there is no separating line between them.
<br />
<br />
<div class='image'>
<img src='polygon_polygon/images/2f.png' />
</div>
<br />
</p>
</section>
<section>
<h2>Algorithm for Finding the Intersection</h2>
<p>
Given two polygons <b>A</b> and <b>B</b>:
<ol>
<li>For each edge on <b>A</b>, get the normal <i>n</i> of that edge.</li>
<li>Project each vertex <i>v</i> of <b>A</b> onto <i>n</i>. Return the minimum and maximum projection of all vertices as the projection of the shape.</li>
<li>Repeat Step 2 for polygon <b>B</b>.</li>
<li>If the min and max projections found in Steps 2 and 3 do <b>NOT</b> overlap, the polygons are not intersecting. Return false.</li>
<li>If the projections overlap for each edge of both shapes, the shapes are intersecting. Return true.</li>
</ol>
And that is all there is to <i>finding</i> the intersection between two convex polygons. Here is the code snippet that does this if you are interested:
<pre><code><span class="code_keyword">Vector2</span> getProjection(Vector2* vertices, <span class="code_keyword">int</span> numVertices, <span class="code_keyword">Vector2</span> axis) {
<span class="code_comment">// Find the min and max vertex projections</span>
<span class="code_keyword">float32</span> min = axis.dot(vertices[0]);
<span class="code_keyword">float32</span> max = min;
for (int v = 1; v < numVertices; v++) {
<span class="code_keyword">float32</span> d = axis.dot(vertices[v]);
if (d < min) {
min = d;
} else if (d > max) {
max = d;
}
}
<span class="code_keyword">return</span> <span class="code_keyword">Vector2</span> { min, max };
}
<span class="code_keyword">bool</span> projectionsOverlap(Vector2 first, <span class="code_keyword">Vector2</span> second) {
<span class="code_keyword">return</span> first.x <= second.y && second.x <= first.y;
}
<span class="code_keyword">bool</span> doIntersect(ConvexPolygon* first, ConvexPolygon* second) {
IntersectionResult ir;
<span class="code_comment">// Check agaisnt the edges of the first polygon</span>
for (int i = 0; i < first->numVertices; i++) {
<span class="code_keyword">Vector2</span> normal = first->edges[i].normal;
<span class="code_keyword">Vector2</span> firstProj = getProjection(first->transformedVertices, first->numVertices, normal);
<span class="code_keyword">Vector2</span> secondProj = getProjection(second->transformedVertices, second->numVertices, normal);
if (!projectionsOverlap(firstProj, secondProj)) {
<span class="code_keyword">return</span> false;
}
}
<span class="code_comment">// Check against the edges of the second polygon</span>
for (int i = 0; i < second->numVertices; i++) {
<span class="code_keyword">Vector2</span> normal = second->edges[i].normal;
<span class="code_keyword">Vector2</span> firstProj = getProjection(first->transformedVertices, first->numVertices, normal);
<span class="code_keyword">Vector2</span> secondProj = getProjection(second->transformedVertices, second->numVertices, normal);
if (!projectionsOverlap(firstProj, secondProj)) {
<span class="code_keyword">return</span> false;
}
}
<span class="code_keyword">return</span> true;
}
</code></pre> </p>
</section>
<section>
<h2>SAT Collision Resolution</h2>
<p>
Now that we know that our objects have intersected, we want to send them tumbling away from one another in order to simulate a collision. To do this, we will need to find the following things:
<ul>
<li><b>Point of Application</b>: at what point on each object did the objects first intersect</li>
<li><b>Collision Normal</b>: in what direction, point towards object <b>A</b>, did the polygons intersect</li>
<li><b>Relative Velocity</b>: easily found by taking the difference between the two velocities.
</ul>
To find these values, we must first find both the shape and the edge that caused the intersection in the first place. To do this, we will think about what we know so far and try to arrive at some intutitive understanding of it. Keep in mind that I am not a proof-minded person, so you will not be finding that here.
<br/>
<h3>Finding the Intersecting Edge</h3>
We can already figure out that the following two triangles intersect one another:
<br />
<br />
<div class='image'>
<img src='polygon_polygon/images/3a.png' />
</div>
<br/><br/>
We know that <b>A</b> can only intersect <b>B</b> if: (1) a vertex from <b>A</b> is inside of <b>B</b>, (2) an edge of <b>A</b> flatly intersects an edge of <b>B</b>, or (3) a vertex of <b>A</b> overlaps exactly a vertex of <b>B</b>. Honestly, for our purposes, scenarios 2 and 3 are quite unlikely, but we can explore them a bit just to see how we might resolve them. We will start with the first case, since it is more likely. We will start by drawing the axis defined by the leftmost edge of <b>B</b> with both polygons projected onto it:
<br /> <br />
<div class='image'>
<img src='polygon_polygon/images/3c.png' />
</div>
<br/><br/>
This is a poorly drawn picture, but you should be able to see that the bit in green represents the intersection between the projections of the two polygons. If we were to repeat this same exercise for every edge here, we'll begin to see something very interesting. And, if we take new shapes and continue this stategy, we can begin to come to a very elegant conclusion:
<br/>
<br/>
<i>The intersecting edge will be the one where the projection of triangle <b>A</b> overlaps with the projection of triangle <b>B</b> the <b>least</b></i>!
<br /> <br />
I'm sure someone more inclined to proving mathematical truities would love to describe this to you, but, for all intents and purposes, this intutitive understanding is good enough for us. We just want to make games, anyhow.
<br />
<h3>Finding which point causes the intersection</h3>
</p>
</section>
<section>
<h2>
Live Example of Intersection Detection
</h2>
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<h2>References</h2>
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