Figure 1: Two projections of one vector onto another. Figure 2: Projecting a non-unit vector.

Transcription

1 w w Figure 1: Two projections of one ector onto another. w Figure 2: Projecting a non-unit ector. 1 Intersecting a Ray with a Triangle But we don t care about whether the ray intersects the plane, we care whether it intersects a triangle or a quad. We can compute the point of intersection (from the parameter and our parametric equation of the ray), and then try to compute the s and t alues so that we can use the constraints in section??. 1.1 Dot Products as Projections Before we deelop the code for finding the intersection point within a triangle, it s helpful to hae as a building block a more complete understanding the usefulness of the dot product. We know that the dot product gies us something like the cosine of the angle between two ectors. In fact, for unit ectors, it gies us exactly the cosine of the angle between them. Let s start with unit ectors. Suppose we hae unit ectors and w. If we compute the following: = ( w) w = ( w)w The and w ectors are scalings of the original ectors, where they are scaled by the dot product. Geometrically, this is equialent to the projection of the other ector onto this one; see figure 1. 1

2 u u w t a b Figure 3: Taking the dot product of w with u (shown as u ). We start with known ectors u, and w, where w is some linear combination of u and : w = su + t. We first find u, the perpendicular to u lying in the plane (denoted u in the figure). Taking the dot product finds t, the amount of ector that is in the linear combination of u and to make up w. But what about non-unit ectors? Consider the projection in figure 2 and assume that is a unit ector but w is not. To project w onto, we want to resize so that it has a length equal to the length of w multiplied by the cosine of the angle between and w. Call that angle θ. The length of should be: Since = 1, we can drop that, and we find that: w cos(θ) = w w w = w The dot product gies the projection of any ector onto a unit ector. 1.2 Finding the Intersection Point Parameters If I is the intersection point of the line and the plane, we hae: I = P 0 + s(p 1 P 0 ) + t(p 2 P 0 ) = P 0 + su + t We hae to sole this for s and t, the parameters of the intersection point. Since we hae three dimensions in two unknowns, we can certainly sole this. Indeed, we can sole it three different ways, depending on which equation we decide to leae out. I consulted GeometryAlgorithms.com and decided, for better or worse, to sole it using their math. Here s my explanation of their math. If you d like to see their way, consult algorithm0105. Let w be the ector from P 0 to the intersection point: w = I P 0. We want to sole the following equation for s and t. w = su + t Note that this equation just says that w is a linear combination of ectors u and. They sole this in a ery cleer way. To sole for t, they construct a ector that is orthogonal (perpendicular) to u but that also lies in the plane; call it u, pronounced u perp. The dot product of u with u is, of course, zero, so taking the dot product of the right side of this equation with u nullifies the su term, leaing an equation with only the t parameter to sole for. Of course, there are infinitely many ectors perpendicular to u; it s important that they choose one that lies in the plane, since that gies us the situation shown in figure 3. 2

3 In that figure, a and b are the projections of w and onto u. The scalar multiples are: a = w u b = u By similar triangles, the ector t is to as a is to b: t = a b Therefore, we can find t = a/b. Here it is again, purely algebraically: w w u = su + t = (su + t) u = s(u u ) + t( u ) = t( u ) t = w u u Note that, because the numerator and denominator both hae u in them, it doesn t matter whether u is a unit ector, because the scale factor to normalize it appear in both the numerator and denominator and therefore would cancel. Similarly, we can sole for s by finding a that is perpendicular to and lies in the plane t. s = w u How can we find these perpendicular ectors? Since they lie in the plane, they must be perpendicular to the plane normal, N. Since the cross product finds a ector perpendicular to two others, we hae: u = N u = N Next, they introduce a computational shortcut. It turns out that there is an identity for cross products, namely: (a b) c = (a b)b (b c)a We re not een going to think about proing that; we re just going to use it. Consequently, u = n u = (u ) u = (u u) (u )u = n = (u ) = (u ) ( )u And now we can compute s and t using only dot products. s = t = (u )(w ) ( )(w u) (u ) 2 (u u)( ) (u )(w u) (u u)(w u) (u ) 2 (u u)( ) Notice the similarity between the two calculations. Thus, the complete calculation only requires fie distinct dot products. We need to check for special cases where the triangle is degenerate: 3

4 One way this can happen is if two of the points defining the triangle are the same. If this happens, either u or will be the zero ector, and u u = 0 or = 0. Another way is if the three points are colinear, in which case u is a scalar multiple of. One way to test for this is to test whether the cosine of the angle between u and is 1, which happens when the angle is zero. u u = 1 u = u (u ) 2 = (u u)( ) (u ) 2 (u u)( ) = 0 Since the quantity on the left is the denominator of our fractions to compute s and t, all we need to do is check for zero before diiding. 2 Boxing Tests The code aboe (both ours and the code in Three.js) is computationally expensie, so we want to aoid it if we can. Hence, we should use boxing tests to aoid the expensie computation in cases that are easy. The Three.js code uses two boxing tests: it first tests if the line is too far from the triangle by using a sphere that surrounds the triangle. Then, it tests if the line can t intersect the triangle because it doesn t intersect a box surrounding the triangle. If both tests are passed, the expensie test is used. Here s the oerall algorithm (in release 67; it s changed since then): ar intersectobject = function ( object, raycaster, intersects ) { if ( object instanceof THREE.Sprite ) { //... else if ( object instanceof THREE.Mesh ) { ar geometry = object.geometry; // Checking boundingsphere distance to ray if ( geometry.boundingsphere === null ) geometry.computeboundingsphere(); sphere.copy( geometry.boundingsphere ); sphere.applymatrix4( object.matrixworld ); if ( raycaster.ray.isintersectionsphere( sphere ) === false ) { return intersects; // Check boundingbox before continuing inersematrix.getinerse( object.matrixworld ); localray.copy( raycaster.ray ).applymatrix4( inersematrix ); 4

5 if ( geometry.boundingbox!== null ) { if ( localray.isintersectionbox( geometry.boundingbox ) === false ) { return intersects; if ( geometry instanceof THREE.BufferGeometry ) { //... else if ( geometry instanceof THREE.Geometry ) { ar isfacematerial = object.material instanceof THREE.MeshFaceMaterial; ar objectmaterials = isfacematerial === true? object.material.materials : null; ar a, b, c, d; ar precision = raycaster.precision; ar ertices = geometry.ertices; for ( ar f = 0, fl = geometry.faces.length; f < fl; f ++ ) { ar face = geometry.faces[ f ]; ar material = isfacematerial === true? objectmaterials[ face.materialindex ] : object.material; if ( material === undefined ) continue; a = ertices[ face.a ]; b = ertices[ face.b ]; c = ertices[ face.c ]; if ( material.side === THREE.BackSide ) { ar intersectionpoint = localray.intersecttriangle( c, b, a, true ); else { ar intersectionpoint = localray.intersecttriangle( a, b, c, material.side!== THREE.DoubleSide ); if ( intersectionpoint === null ) continue; 5

6 intersectionpoint.applymatrix4( object.matrixworld ); ar distance = raycaster.ray.origin.distanceto( intersectionpoint ); if ( distance < precision distance < raycaster.near distance > raycas intersects.push( { ); distance: distance, point: intersectionpoint, face: face, faceindex: f, object: object Let s look in more detail at the two boxing tests. distance from a point to a line bounding containers Here s the code for isintersectionsphere: isintersectionsphere: function ( sphere ) { return this.distancetopoint( sphere.center ) <= sphere.radius;, and the distancetopoint function is: distancetopoint: function () { ar 1 = new THREE.Vector3(); return function ( point ) { ar directiondistance = 1.subVectors( point, this.origin ).dot( this.direction ); // point behind the ray if ( directiondistance < 0 ) { return this.origin.distanceto( point ); 1.copy( this.direction ).multiplyscalar( directiondistance ).add( this.origin ); 6

7 return 1.distanceTo( point ); ; (), The box intersection test is a lot of code, so I won t include it here. We ll go oer all this in more detail in class, with some example problems. 7