Question

具有一个带有两个或多个相交面的加载网格，我想用不同的颜色显示这些面。我真的不需要交点或线段，我只是在寻找最快的方法来知道两个面/三角形是否相交。在此示例中，有三个相交的面应以红色显示（materialIndex = 1）。

最初，我基于对光线B进行了三次测试（基于A光线投射器三次（针对A面的每个边缘，将光线投射器限制在两点之间的距离）），但是我无法使其正常工作（有是在远/错的地方检测到大量错误的交叉路口。

TomasMöller的方法似乎是检测这些交叉点的较快方法之一。我试图将涉及的计算迁移到Three.JS，我认为一切都在正确的地方，但是结果不是预期的。

/**
 * -----------------------------------------------------------------------
 * "A Fast Triangle-Triangle Intersection Test" (by TOMAS MOLLER)
 * http://web.stanford.edu/class/cs277/resources/papers/Moller1997b.pdf
 * -----------------------------------------------------------------------
 * 
 * Let us denote the two triangles T1 and T2; the vertices of T1 and T2
 * by V10,V11,V12, and V20,V21,V22 respectively; and the planes in which the
 * triangles lie #1 and #2. First, the plane equation #2: N2 · X + d2 = 0 
 * (where X is any point on the plane) is computed: (1)
 * 
 *      N2 = (V2.1 - V2.0) * (V2.2 - V2.0)
 *      d2 = -N2 · V2.0
 * 
 * Then the signed distances from the vertices of T1 to #2 (multiplied by 
 * a constant N2 · N2) are computed by simply inserting the vertices 
 * into the plane equation: (2)
 * 
 *      dV1.i = N2 · V1.i + d2        (i=0,1,2)
 * 
 * Now, if all dV1.i != 0   (i=0,1,2) (that is, no point is on the plane) 
 * and all have the same sign, then T1 lies on one side of #2 and the overlap 
 * is rejected. The same is done for T2 and #1. These two early rejection tests 
 * avoid a lot of computations for some triangle pairs. Indeed, for a pair to 
 * pass this test there must be some line of direction N1 * N2 that meets both.
 * 
 * If all dV1.i = 0   (i=0,1,2) then the triangles are co-planar, and this case 
 * is handled separately. If not, the intersection of #1 and #2 is aline, 
 * L = O + tD, where D = N1 * N2 is the direction of the line and O is some point 
 * on it. Note that due to our previous calculations and rejections, both triangles 
 * are guaranteed to intersect L. These intersections form intervals on L, and 
 * if these intervals overlap, the triangles overlap as well.
 **/

此JSFiddle包含我正在使用的迁移代码。我正在使用归一化向量，但我想我缺少使用这些值进行的最终计算。

function checkIntersectedFaces(){
    var face, vA, vB, vC, vAnorm, vBnorm, vCnorm;
    var face2, vA2, vB2, vC2, vA2norm, vB2norm, vC2norm;   

    // We only focus on the small stand-alone triangle
    console.group("CHECKING MAIN FACE (#4)");
    face = faces[4];    

    vA = vertices[face.a].clone();
    vB = vertices[face.b].clone();
    vC = vertices[face.c].clone();
    console.groupCollapsed("face vertices");
    console.log("vA, vB, vC");
    console.log([vA, vB, vC]);
    console.groupEnd("face vertices");

    vAnorm = vA.clone().normalize();
    vBnorm = vB.clone().normalize();
    vCnorm = vC.clone().normalize();
    console.groupCollapsed("face vertices (normalized)");
    console.log("vAnorm, vBnorm, vCnorm");
    console.log([vAnorm, vBnorm, vCnorm]);
    console.groupEnd("face vertices (normalized)");

    // Compare main face (stand-alone triangle) against all other faces
    for(var faceIndex=0; faceIndex< faces.length; faceIndex++){
        if(faceIndex == 4) continue; // avoid self-comparison
        console.group("VS FACE #" + faceIndex); 
        face2 = faces[faceIndex];     
        if(faceIndex == 1 || faceIndex == 3) 
            console.warn("This face should be detected as INTERSECTED");

        vA2 = vertices[face2.a].clone();
        vB2 = vertices[face2.b].clone();
        vC2 = vertices[face2.c].clone();
        console.groupCollapsed("face vertices");
        console.log("vA2, vB2, vC2");
        console.log([vA2, vB2, vC2]);
        console.groupEnd("face vertices");

        vA2norm = vA2.clone().normalize();
        vB2norm = vB2.clone().normalize();
        vC2norm = vC2.clone().normalize();
        console.groupCollapsed("face vertices (normalized)");
        console.log("vA2norm, vB2norm, vC2norm");
        console.log([vA2norm, vB2norm, vC2norm]);
        console.groupEnd("face vertices (normalized)");


        /////////////////////////////////////////////////////////////
        // MOLLER calculations
        /////////////////////////////////////////////////////////////
        var n2 = vB2.sub(vA2) .multiply( vC2.sub(vA2) );
        var d2 = -n2.dot( vA2 );
        var dA = n2.dot( vA ) + d2;
        var dB = n2.dot( vB ) + d2;
        var dC = n2.dot( vC ) + d2;

        console.groupCollapsed("n2");
        console.log(n2);
        console.log("d2", d2);
        console.log("dA", dA);
        console.log("dB", dB);
        console.log("dC", dC);
        console.groupEnd("n2");
        /////////////////////////////////////////////////////////////
        var n2norm = vB2norm.sub(vA2norm) .multiply( vC2norm.sub(vA2norm) );
        var d2norm = -n2norm.dot( vA2norm );
        var dAnorm = n2norm.dot( vAnorm ) + d2norm;
        var dBnorm = n2norm.dot( vBnorm ) + d2norm;
        var dCnorm = n2norm.dot( vCnorm ) + d2norm; 

        console.groupCollapsed("n2 (normalized)");
        console.log(n2norm);
        console.log("d2", d2norm);
        console.log("dA", dAnorm);
        console.log("dB", dBnorm);
        console.log("dC", dCnorm);
        console.groupEnd("n2 (normalized)");

        /////////////////////////////////////////////////////////////
        // CHECK INTERSECTIONS
        /////////////////////////////////////////////////////////////
        if(dAnorm!=0 && dBnorm!=0 && dCnorm!=0){
            console.warn("NO INTERSECTION");
        }
        else if(dAnorm==0 && dBnorm==0 && dCnorm==0){
            console.warn("CO-PLANAR FACES");
        }
        else{
            console.warn("INTERSECTION FOUND!");
            face.materialIndex = 1;
            face2.materialIndex = 1;
        }     
        /////////////////////////////////////////////////////////////
        /////////////////////////////////////////////////////////////   

        console.groupEnd("VS FACE #" + faceIndex);
    }
    console.groupEnd("CHECKING MAIN FACE (#4)");
}

Answer 1

我在this thread中找到了一个可行的解决方案，建议使用Ray.js及其方法 intersectTriangle（）（而不是Raycaster）。如果您要比较每个面的三个顶点，则它更快且易于使用。然后，您可以将主面的3个边缘中的每一个相对于其他面进行光线投射。它返回一个向量，该向量具有边与面的交点。

但是似乎您必须遵循某些步骤才能避免在发射光线时出错。首先，将输出变量定义为Vector3。然后，将intersectTriangle（）调用的结果分配给该变量，并且还将此输出变量用作调用中的目标。像这样的东西，它看起来很多余，但无法通过其他方式起作用：

var ray = new THREE.Ray();
    ray.set(sourceVector, directionVector);
var intersect = new THREE.Vector3();
    intersect = ray.intersectTriangle(triangle[0], triangle[1], triangle[2], false, intersect);

（“三角形”是一个简单的数组，其中包含构成要比较的面部的3个向量）

Three.JS中的三角形与三角形的交点

1 个答案: