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Monday, April 4, 2011

2D Polygon - Raycasting

Ever write a 2D game that supported convex polygons? Ever struggle to find good sources for ray casting onto polygons in 2D space?

Ironically, in 3D you'll find a number of different sources for Ray-Polygon intersection. Unless BSP trees are in use it usually involves some procedure resembling the following:

1) Derive a plane for the polygon, generally using the surface normal.
2) Find the intersection point of the ray and the plane (if any)
3) See if the polygon contains the point. There are a number of ways to do this, most notably using a binary search and seeing if the new point constructs a CCW triangle.

In 2D, the test can't really work this way. Instead we need to test the ray against each segment of the polygon. You might assume that if any intersection with a segment is found, we can conclude there is a valid intersection. However, this is not the case. There are two reasons we can't do an "early out" test in which the algorithm terminates with a true result if just one of the segments is intersected by the ray.

1) If a polygon contains the ray origin, there is no intersection. A usual PolygonContainsPoint() implementation can serve as a reasonably fast prerequisite test. However, it would be faster if we used the ray casting algorithm. If you read up on the algorithm you'll see it involves casting onto every segment and counting the number of crossings. With an even number, we can conclude the origin is outside the polygon, otherwise it resides inside of it.
2) When ray casting against a polygon we'll usually want far more information than just a boolean result. We'll usually want the distance of the ray, the contact point, and the normal of the ray intersection. In all these cases, we'll always want the closest intersection point. To definitively find it we'll need to test against every segment anyways. Running a contains method won't result in twice as many iterations (since it uses a binary search), but it will result in more than just one sweep of the polygon.

Ray-Segment Test
The ray segment test in 2D can be solved parametrically, as explained here. First we check if the ray direction and the segment are parallel, in which case we can quickly cancel out the possibility of a collision*. Then we solve for s and t. S must be a value between 0 and 1 to lie on the segment. T must be between zero (or some other user defined min value) and the max t value.

* This is false for collinear segments and rays. I have not found a clean and efficient way to deal with this case. If you have a suggestion, comment below. Using distance checks will work, but is an ugly way to check if a point is within the range of tmax. The normal in this case would probably have to be the ray direction reflected off the vertex normal.

And of course, the code for the test. Please note that LeftPerp(v) = (v.Y, -v.X). The special equals call simply checks if the value is within epsilon distance from zero. Please note that Epsilon is rarely the most appropriate tolerance value, but what is varies widely depending on your units and scaling and I found it was sufficient towards understanding the algorithm itself.  Still: numerical robustness matters!

public static bool RayIntersectsSegment(Ray ray, Vector2 pt0, Vector2 pt1, float tmax, out float t) {
    Vector2 seg = pt1 - pt0;
    Vector2 segPerp = LeftPerp(seg);
    float perpDotd = Vector2.Dot(ray.Direction, segPerp);
    if (Equals(perpDotd, 0.0f, float.Epsilon))
        t = float.MaxValue;
        return false;

    Vector2 d = pt0 - ray.Origin;

    t = Vector2.Dot(segPerp, d) / perpDotd;
    float s = Vector2.Dot(LeftPerp(ray.Direction), d) / perpDotd;

    return t >= 0.0f && t <= tmax && s >= 0.0f && s <= 1.0f;
Ray-Polygon Test
The final ray polygon test brings all of this together in a relatively simple way. As we iterate through the polygon we test every edge. If there is an intersection we increment the number of crossings. If the distance is less than the current minimum distance we now have a new closest intersection, and we proceed to calculate the normals and contact point for this edge. If the number of crossings is even and greater than zero at the end of the test, a true result is returned and the current distance, contact point, and normal are used.
Contact point shown in light green, ray reflection normal in purple
Update: The purple line above is actually drawing the reflection of the ray off the normal, not the edge normal itself.  The below code has been corrected to return the edge normal.  In practice, it is much more useful than the reflection (which can always be calculated afterwards using the normal if you need it).

public static bool RayCast(Ray ray, Polygon polygon, float tmax, out float t, out Vector2 pt, out Vector2 normal)
            t = float.MaxValue;
            pt = ray.Origin;
            normal = ray.Direction;
            // temp holder for segment distance
            float distance;
            int crossings = 0;

            for (int j = polygon.NumVertices - 1, i = 0; i < polygon.NumVertices; j = i, i++)
                if (RayIntersectsSegment(ray, polygon.v[j], polygon.v[i], float.MaxValue, out distance))
                    if (distance < t && distance <= tmax)
                        t = distance;
                        pt = ray.GetPoint(t);

                        Vector2 edge = polygon.v[j] - polygon.v[i];
                        // We would use LeftPerp() if the polygon was
                        // in clock wise order
                        normal = Vector2.Normalize(RightPerp(edge));
            return crossings > 0 && crossings % 2 == 0;
In this implementation, it is assumed that the polygon is a convex array of vertices in Counter Clockwise Order. This is why we reflect off the right perpendicular of the edge. If your implementation uses Clockwise polygons you'll want to use LeftPerp() instead. Here's a reference of the helper functions:
public static Vector2 LeftPerp(Vector2 v)
            return new Vector2(v.Y, -v.X);

        public static Vector2 RightPerp(Vector2 v)
            return new Vector2(-v.Y, v.X);

Happy raycasting!

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