这是我在stackoverflow上的第一个问题，所以我希望我把所有事情都做对了。对于java项目，我需要使用八叉树进行光线跟踪。我已经创建了一个简单的八叉树（没有邻居信息或其他东西），并将对象网格的三角形分类到八叉树的AABB中。现在我想通过树轻松遍历每条光线。 （它应该非常简单，因为完成这个项目的时间非常短）。基本算法如下：
不幸的是，我目前的实施似乎有一个＆＃34;错误＆＃34;在交叉点计算中（ray vs ABBB）。我检查AABB的任何一侧是否被击中并记住最近的ip（距射线原点的最小距离）。
以下是我的BoundingBox类中此函数的代码：
public HitResult intersects6(Ray ray) {
double t;
Vec3d ip = new Vec3d();
HitResult finalHitResult = null;
// front xy
if (Math.abs(ray.direction.z) > Helper.EPSYLON) {
t = (vmax.z - ray.origin.z) / ray.direction.z;
ip.x = ray.origin.x + t * ray.direction.x;
ip.y = ray.origin.y + t * ray.direction.y;
ip.z = vmax.z;
if ((ip.x >= vmin.x) && (ip.x <= vmax.x) && (ip.y >= vmin.y) && (ip.y <= vmax.y)) {
// here is an intersection
double distance = Vec3d.distance(ray.origin, ip);
finalHitResult = new HitResult(ip, distance);
}
}
// back xy
if (Math.abs(ray.direction.z) > Helper.EPSYLON) {
t = (vmin.z + ray.origin.z) / -ray.direction.z;
ip.x = ray.origin.x + t * ray.direction.x;
ip.y = ray.origin.y + t * ray.direction.y;
ip.z = vmin.z;
if ((ip.x >= vmin.x) && (ip.x <= vmax.x) && (ip.y >= vmin.y) && (ip.y <= vmax.y)) {
double distance = Vec3d.distance(ray.origin, ip);
if (finalHitResult!= null) {
if(distance < finalHitResult.distance)
finalHitResult.distance = distance;
finalHitResult.point = ip;
}
else
finalHitResult = new HitResult(ip, distance);
}
}
// Side Right
if (Math.abs(ray.direction.x) > Helper.EPSYLON) {
t = (vmax.x - ray.origin.x) / ray.direction.x;
ip.y = ray.origin.y + t * ray.direction.y;
ip.z = ray.origin.z + t * ray.direction.z;
ip.x = vmax.x;
if ((ip.y >= vmin.y) && (ip.y <= vmax.y) && (ip.z >= vmin.z) && (ip.z <= vmax.z)) {
double distance = Vec3d.distance(ray.origin, ip);
if (finalHitResult!= null) {
if(distance < finalHitResult.distance)
finalHitResult.distance = distance;
finalHitResult.point = ip;
}
else
finalHitResult = new HitResult(ip, distance);
}
}
// Side Left
if (Math.abs(ray.direction.x) > Helper.EPSYLON) {
t = (vmin.x + ray.origin.x) / -ray.direction.x;
ip.y = ray.origin.y + t * ray.direction.y;
ip.z = ray.origin.z + t * ray.direction.z;
ip.x = vmin.x;
if ((ip.y >= vmin.y) && (ip.y <= vmax.y) && (ip.z >= vmin.z) && (ip.z <= vmax.z)) {
double distance = Vec3d.distance(ray.origin, ip);
if (finalHitResult!= null) {
if(distance < finalHitResult.distance)
finalHitResult.distance = distance;
finalHitResult.point = ip;
}
else
finalHitResult = new HitResult(ip, distance);
}
}
// Top
if (Math.abs(ray.direction.y) > Helper.EPSYLON) {
t = (vmax.y - ray.origin.y) / ray.direction.y;
ip.x = ray.origin.x + t * ray.direction.x;
ip.z = ray.origin.z + t * ray.direction.z;
ip.y = vmax.y;
if ((ip.x >= vmin.x) && (ip.x <= vmax.x) && (ip.z >= vmin.z) && (ip.z <= vmax.z)) {
double distance = Vec3d.distance(ray.origin, ip);
if (finalHitResult!= null) {
if(distance < finalHitResult.distance)
finalHitResult.distance = distance;
finalHitResult.point = ip;
}
else
finalHitResult = new HitResult(ip, distance);
}
}
// Bottom
if (Math.abs(ray.direction.y) > Helper.EPSYLON) {
t = (vmin.y + ray.origin.y) / -ray.direction.y;
ip.x = ray.origin.x + t * ray.direction.x;
ip.z = ray.origin.z + t * ray.direction.z;
ip.y = vmin.y;
if ((ip.x >= vmin.x) && (ip.x <= vmax.x) && (ip.z >= vmin.z) && (ip.z <= vmax.z)) {
double distance = Vec3d.distance(ray.origin, ip);
if (finalHitResult!= null) {
if(distance < finalHitResult.distance)
finalHitResult.distance = distance;
finalHitResult.point = ip;
}
else
finalHitResult = new HitResult(ip, distance);
}
}
return finalHitResult;
我想这不是最好的方法。在我的第一个实现中，我只使用了t值并对它们进行了比较（找到我想要访问的框）。但问题是一样的。找不到一些交叉点。
我还在这里查看了交集方法： https://code.google.com/p/3d-workspace/source/browse/trunk/MathLibrary/Bounding/BoundingBox.cpp?r=17 但我无法看到如何获得与此代码（甚至任何t值）的交叉点。此外，我测试了slab方法，如下所示： http://tavianator.com/2011/05/fast-branchless-raybounding-box-intersections/ 但这似乎也错过了一些交叉路口，我不知道为什么：
public double[] intersects3(Ray ray) {
double Tnear = -1e30;
double Tfar = 1e30;
// First, check slab in X.
if (Math.abs(ray.direction.x) < 0.0) {
// Ray is parallel to X, but starts outside. Fail.
if (ray.origin.x < vmin.x || ray.origin.x > vmax.x) {
return null;
}
} else {
double Ta = ((vmin.x - ray.origin.x) / ray.direction.x), Tb = (vmax.x - ray.origin.x) / ray.direction.x;
double T1 = Math.min(Ta, Tb);
double T2 = Math.max(Ta, Tb);
if (T1 > Tnear)
Tnear = T1;
if (T2 < Tfar)
Tfar = T2;
if (Tnear > Tfar)
return null;
if (Tfar < 0)
return null;
}
// Then check slab in Y.
if (Math.abs(ray.direction.y) < 0.0) {
// Ray is parallel to X, but starts outside. Fail.
if (ray.origin.y < vmin.y || ray.origin.y > vmax.y) {
return null;
}
} else {
double Ta = (vmin.y - ray.origin.y) / ray.direction.y, Tb = (vmax.y - ray.origin.y) / ray.direction.y;
double T1 = Math.min(Ta, Tb);
double T2 = Math.max(Ta, Tb);
if (T1 > Tnear)
Tnear = T1;
if (T2 < Tfar)
Tfar = T2;
if (Tnear > Tfar)
return null;
if (Tfar < 0)
return null;
}
// Then check slab in Z.
if (Math.abs(ray.direction.z) < 0.0) {
// Ray is parallel to X, but starts outside. Fail.
if (ray.origin.z < vmin.z || ray.origin.z > vmax.z) {
return null;
}
} else {
double Ta = (vmin.z - ray.origin.z) / ray.direction.z, Tb = (vmax.z - ray.origin.z) / ray.direction.z;
double T1 = Math.min(Ta, Tb);
double T2 = Math.max(Ta, Tb);
if (T1 > Tnear)
Tnear = T1;
if (T2 < Tfar)
Tfar = T2;
if (Tnear > Tfar)
return null;
if (Tfar < 0)
return null;
}
// If we have survived this far, the test passed.
return new double[] { Tnear, Tfar };
}
也许我对光线追踪太愚蠢了。
但我的实际问题是：
是否可以使用t值来比较哪个框具有最近的交叉点？如果是的话，我怎样才能得到这个t值？或者我该怎么做才能使第一个代码剪断工作？ （到目前为止，我对任何工作解决方案都很满意，即使这个解决方案非常慢）
提前致谢。
答案 0 :(得分：-1)
也许这可能会有所帮助： http://chiranjivi.tripod.com/octrav.html
我试图实现四叉树的想法： https://github.com/alexroat/quadtree-traversal