使用THREE.Frustum计算近/远平面顶点

Question

我需要一些帮助来处理THREE.Frustum对象.

我的问题:

我需要计算近/远平面顶点; 我看过这些教程

1. http://www.lighthouse3d.com/tutorials/view-frustum-culling/view-frustums-shape/
2. http://www.lighthouse3d.com/tutorials/view-frustum-culling/geometric-approach-extracting-the-planes/

并且我已经勾勒出这个函数完全实现(我希望如此)所解释的过程(只是为了得到左上/右上顶点,假设相机只能左右看):

        // Near Plane dimensions
        hNear = 2 * Math.tan(camera.fov / 2) * camera.near; // height
        wNear = hNear * camera.aspect; // width

        // Far Plane dimensions
        hFar = 2 * Math.tan(camera.fov / 2) * camera.far; // height
        wFar = hFar * camera.aspect; // width

getVertices : function() {
        var p = camera.position.clone();
        var l = getCurrentTarget(); // see below
        var u = new THREE.Vector3(0, 1, 0);

        var d = new THREE.Vector3();
        d.sub(l, p);
        d.normalize();

        var r = new THREE.Vector3();
        r.cross(u, d);
        r.normalize();

        // Near Plane center
        var dTmp = d.clone();
        var nc = new THREE.Vector3();
        nc.add(p, dTmp.multiplyScalar(camera.near));

        // Near Plane top-right and top-left vertices
        var uTmp = u.clone();
        var rTmp = r.clone();
        var ntr = new THREE.Vector3();
        ntr.add(nc, uTmp.multiplyScalar(hNear / 2));
        ntr.subSelf(rTmp.multiplyScalar(wNear / 2));

        uTmp.copy(u);
        rTmp.copy(r);
        var ntl = new THREE.Vector3();
        ntl.add(nc, uTmp.multiplyScalar(hNear / 2));
        ntl.addSelf(rTmp.multiplyScalar(wNear / 2));

        // Far Plane center
        dTmp.copy(d);
        var fc = new THREE.Vector3();
        fc.add(p, dTmp.multiplyScalar(camera.far));

        // Far Plane top-right and top-left vertices
        uTmp.copy(u);
        rTmp.copy(r);
        var ftr = new THREE.Vector3();
        ftr.add(fc, uTmp.multiplyScalar(hFar / 2));
        ftr.subSelf(rTmp.multiplyScalar(wFar / 2));

        uTmp.copy(u);
        rTmp.copy(r);
        var ftl = new THREE.Vector3();
        ftl.add(fc, uTmp.multiplyScalar(hFar / 2));
        ftl.addSelf(rTmp.multiplyScalar(wFar / 2));

getCurrentTarget : function() {
        var l = new THREE.Vector3(0, 0, -100);
        this.camera.updateMatrixWorld();
        this.camera.matrixWorld.multiplyVector3(l);
        return l;
    }

这似乎有效,但......

我的问题:

使用THREE.Frustum对象,我可以以更优雅(可能更正确)的方式获得相同的结果吗？

Answer 1

Three.Frustum并不是真的会帮助你 - 它是一组飞机.好消息是你的解决方案看似正确,但有一种更简单的方法可以考虑这一点.

近平面的右上角是相机空间中具有以下坐标的点:

var ntr = new THREE.Vector3( wNear / 2, hNear / 2, -camera.near );

使用你的定义wNear和hNear,这是正确的.

现在,确保camera.matrixWorld更新,您将该点转换为世界坐标,如下所示:

camera.updateMatrixWorld();
ntr.applyMatrix4( camera.matrixWorld );

现在,翻转标志以获得其他三个角,然后重复远平面的计算.

看,你做对了; 你刚刚走了一条更复杂的路线.:-)

编辑:更新为three.js r.66