Java 这个光线投射算法正确吗？这是测试直线/光线与立方体相交的正确方法吗？

我正在建立一个LWJGL程序，测试立方体是否在摄像机前面。我认为这是可行的，但我得到了大量不同的数据。我认为这是因为平面永远在运动，它们不是有限的正方形。我该怎么做才能解决这个问题？（我也注意到我的代码相当慢，是什么原因造成的？）

这是我的交叉口代码：

public static float getCubeInteresection(Cube b, Vector3f camera, Vector3f lookat)
{
    Vector3f l = b.Location;
    float boxs = 0.5f;
    Plane3f[] bplanes = Plane3f.getBoxDefaults(l, boxs);
    Vector3f result = null;
    for (int t = 0; t < bplanes.length; t++)
    {
        Vector3f raytrace = getIntersection(camera, lookat,bplanes[t].Point0,bplanes[t].Point1,bplanes[t].Point2);
        if (raytrace != null)
        {
            result = raytrace;
            break;
        }
    }
    if (result == null)
        return -1.0f; //not hit, waste of time... lol
    float x = (float) Math.pow(result.x - camera.x, 2);
    float y = (float) Math.pow(result.y - camera.y, 2);
    float z = (float) Math.pow(result.z - camera.z, 2);
    float dist = (float) Math.sqrt(x + y + z);
    return dist;
}
private static Vector3f getIntersection(Vector3f line12, Vector3f line22,Vector3f plane12, Vector3f plane22, Vector3f plane32)
{
    javax.vecmath.Vector3f plane1 = new javax.vecmath.Vector3f(plane12.x,plane12.y,plane12.z);
    javax.vecmath.Vector3f plane2 = new javax.vecmath.Vector3f(plane22.x,plane22.y,plane22.z);
    javax.vecmath.Vector3f plane3 = new javax.vecmath.Vector3f(plane32.x,plane32.y,plane32.z);
    javax.vecmath.Vector3f line1 = new javax.vecmath.Vector3f(line12.x,line12.y,line12.z);
    javax.vecmath.Vector3f line2 = new javax.vecmath.Vector3f(line22.x,line22.y,line22.z);
    javax.vecmath.Vector3f p1 = new javax.vecmath.Vector3f(plane1);
    javax.vecmath.Vector3f p2 = new javax.vecmath.Vector3f(plane2);
    javax.vecmath.Vector3f p3 = new javax.vecmath.Vector3f(plane3);
    javax.vecmath.Vector3f p2minusp1 = new javax.vecmath.Vector3f(p2);
    p2minusp1.sub(p1);
    javax.vecmath.Vector3f p3minusp1 = new javax.vecmath.Vector3f(p3);
    p3minusp1.sub(p1);
    javax.vecmath.Vector3f normal = new javax.vecmath.Vector3f();
    normal.cross(p2minusp1, p3minusp1);
    double d = -p1.dot(normal);
    javax.vecmath.Vector3f i1 = new javax.vecmath.Vector3f(line1);
    javax.vecmath.Vector3f direction = new javax.vecmath.Vector3f(line1);
    direction.sub(line2);
    double dot = direction.dot(normal);
    if (dot == 0) return null;
    double t = (-d - i1.dot(normal)) / (dot);
    javax.vecmath.Vector3f intersection = new javax.vecmath.Vector3f(line1);
    javax.vecmath.Vector3f scaledDirection = new javax.vecmath.Vector3f(direction);
    float scalent = (float)t;
    scaledDirection.scale(scalent);
    intersection.add(scaledDirection);
    javax.vecmath.Vector3f intersectionPoint = new javax.vecmath.Vector3f(intersection);
    return new Vector3f(intersectionPoint.x,intersectionPoint.y,intersectionPoint.z);
}

控制台的输出：

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摄影机：（0.0,0.0,6.0）注视（线的末端）：（124.33239274.9127，-1000.5496）距离（从“GetBlockInterestment”返回）：2.6099026立方体位置：6,7,8

我必须在游戏中处理这个问题，但是光线投射和截锥体剔除是两件不同的事情。在大量对象上执行截头体剔除通常成本较高，精确度也较低，但它更简单，可能适合您的程序

如果只需要知道立方体是否在视口中，可以使用链接中的函数。这将测试立方体是否在视锥体内，而不仅仅是在它前面，因此如果它在前面但离侧面太远，它将返回false

否则，您将需要使用新的glortho和gldeproject编写光线投射算法

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你说的“我得到了很多不同的数字”是什么意思？您的输出仅显示一个交叉点。抱歉，我的意思是，每次场景更新时，我都会得到不同的数字（数字表示从getBlockIntersection返回的距离），我会移动摄影机的注视点，但实际位置保持静止。摄影机和注视点变量都会传递到交叉点代码，所以我认为，如果你改变其中一个，你会得到不同的数字，这并不奇怪。对，但我应该使用平面，还是应该使用正方形/四边形？如果是这样的话，我该怎么做呢？如果你只是在寻找与框的交点，那么你应该寻找轴对齐的边界框交点或定向边界框交点。有很多书都在谈论这个话题。另外，我们有很多关于这方面的资源，例如：。

import org.lwjgl.util.vector.Vector3f;

public class Plane3f
{
public Vector3f Point0;
public Vector3f Point1;
public Vector3f Point2;
public Plane3f(Vector3f p0, Vector3f p1, Vector3f p2)
{
    Point0 = p0;
    Point1 = p1;
    Point2 = p2;
}
public static Plane3f[] getBoxDefaults(Vector3f l, float boxs)
{
    Plane3f[] plane = new Plane3f[24];
    Vector3f p0p0 = new Vector3f(-boxs +  l.x, -boxs +  l.y, boxs +  l.z);
    Vector3f p0p1 = new Vector3f(boxs +  l.x, -boxs +  l.y, boxs +  l.z);
    Vector3f p0p2 = new Vector3f(boxs +  l.x, boxs +  l.y, boxs +  l.z);
    //Vector3f p0p3 = new Vector3f(-boxs +  l.x, boxs +  l.y, boxs +  l.z);
    Plane3f p0 = new Plane3f(p0p0, p0p1, p0p2);

    Vector3f p1p0 = new Vector3f(-boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Vector3f p1p1 = new Vector3f(-boxs +  l.x, boxs +  l.y, -boxs +  l.z);
    Vector3f p1p2 = new Vector3f(boxs +  l.x, boxs +  l.y, -boxs +  l.z);
    //Vector3f p1p3 = new Vector3f(boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Plane3f p1 = new Plane3f(p1p0, p1p1, p1p2);

    Vector3f p2p0 = new Vector3f(-boxs +  l.x, boxs +  l.y, -boxs +  l.z);
    Vector3f p2p1 = new Vector3f(-boxs +  l.x, boxs +  l.y, boxs +  l.z);
    Vector3f p2p2 = new Vector3f(boxs +  l.x, boxs +  l.y, boxs +  l.z);
    //Vector3f p2p3 = new Vector3f(boxs +  l.x, boxs +  l.y, -boxs +  l.z);
    Plane3f p2 = new Plane3f(p2p0, p2p1, p2p2);

    Vector3f p3p0 = new Vector3f(-boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Vector3f p3p1 = new Vector3f(boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Vector3f p3p2 = new Vector3f(boxs +  l.x, -boxs +  l.y, boxs +  l.z);
    Plane3f p3 = new Plane3f(p3p0, p3p1, p3p2);

    Vector3f p4p0 = new Vector3f(boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Vector3f p4p1 = new Vector3f(boxs +  l.x, boxs +  l.y, -boxs +  l.z);
    Vector3f p4p2 = new Vector3f(boxs +  l.x, boxs +  l.y, boxs +  l.z);
    Plane3f p4 = new Plane3f(p4p0, p4p1, p4p2);

    Vector3f p5p0 = new Vector3f(-boxs +  l.x, -boxs +  l.y, -boxs +  l.z);
    Vector3f p5p1 = new Vector3f(-boxs +  l.x, -boxs +  l.y, boxs +  l.z);
    Vector3f p5p2 = new Vector3f(-boxs +  l.x, boxs +  l.y, boxs +  l.z);
    Plane3f p5 = new Plane3f(p5p0, p5p1, p5p2);
    plane[0] = p0;
    plane[1] = p1;
    plane[2] = p2;
    plane[3] = p3;
    plane[4] = p4;
    plane[5] = p5;
    return plane;
}