如果有三角形顶点，如何通过三角形内的任意 x,y 坐标找到 z

发布于 2024-10-28 17:24:41 字数 574 浏览 1 评论 0原文

给定顶点 V1 (x1,y1,z1)、V2 (x2,y2,z2)、V3 (x3,y3,z3)一个三角形 T，如果我知道 (x,y) 位于三角形 Tp (x1,y1) 的投影内，我必须通过它的 x,y 坐标找到该点的 z 坐标，（x2，y2），（x3，y3）。

实际上，3D 中的三角形平面是由等式定义的：Ax+By+Cz+D=0，我可以找到 z = (D-Ax-By)/C 问题是 A、B、C、D 在运行时计算成本太高：

A = y1(z2-z3) + y2(z3-z1) + y3(z1-z2)
B = z1(x2-x3) + z2(x3-x1) + z3(x1-x2)
C = x1(y2-y3) + x2(y3-y1) + x3(y1-y2)
D = -x1(y2*z3 – y3*z2) – x2(y3*z1 – y1*z3) – x3 (y1*z2 – y2*z1)

是否可以使用 opengl 着色器来计算 A、B、C、D？是否有寻找平面系数的优化算法？

原文

Given vertices V1 (x1,y1,z1), V2 (x2,y2,z2), V3 (x3,y3,z3) of a triangle T, I have to find z coordinate of a point by it's x,y coordinate if I know that (x,y) lies within projection of triangle Tp (x1,y1), (x2,y2), (x3,y3).

Actually, triangle plane in 3D is defined by equation: Ax+By+Cz+D=0, and I can find z = (D-Ax-By)/C
The problem is that A, B, C, D are too expensive to calculate in run-time:

A = y1(z2-z3) + y2(z3-z1) + y3(z1-z2)
B = z1(x2-x3) + z2(x3-x1) + z3(x1-x2)
C = x1(y2-y3) + x2(y3-y1) + x3(y1-y2)
D = -x1(y2*z3 – y3*z2) – x2(y3*z1 – y1*z3) – x3 (y1*z2 – y2*z1)

Is it possible to calculate A, B, C, D using, say, opengl shaders? Are there optimized algorithms to find plane coefficients?

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暮年慕年 2024-11-04 17:24:41

该技术称为重心坐标，但维基页面很难理解 -
请参阅 http://www.alecjacobson.com/weblog/?p=1596

float calcY(vec3 p1, vec3 p2, vec3 p3, float x, float z) {
        float det = (p2.z - p3.z) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.z - p3.z);

        float l1 = ((p2.z - p3.z) * (x - p3.x) + (p3.x - p2.x) * (z - p3.z)) / det;
        float l2 = ((p3.z - p1.z) * (x - p3.x) + (p1.x - p3.x) * (z - p3.z)) / det;
        float l3 = 1.0f - l1 - l2;

        return l1 * p1.y + l2 * p2.y + l3 * p3.y;
}

代码来自 http://www.gamedev.net/topic/597393-getting-the-height-of-a-point-on-a-triangle/ - 仔细考虑计算机图形与数学的使用YZ

ps。我不知道有任何使用着色器的更快版本。一种快速的肮脏+解决方案是使用基于顶点高度的颜色渲染三角形，并在 X,Y 处选择像素颜色 - 实际上，这在台式机上永远不会更快，不知道 opengl -es

The technique is called Barycentric coordinates but the wiki page is pretty hard to follow -
See http://www.alecjacobson.com/weblog/?p=1596

float calcY(vec3 p1, vec3 p2, vec3 p3, float x, float z) {
        float det = (p2.z - p3.z) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.z - p3.z);

        float l1 = ((p2.z - p3.z) * (x - p3.x) + (p3.x - p2.x) * (z - p3.z)) / det;
        float l2 = ((p3.z - p1.z) * (x - p3.x) + (p1.x - p3.x) * (z - p3.z)) / det;
        float l3 = 1.0f - l1 - l2;

        return l1 * p1.y + l2 * p2.y + l3 * p3.y;
}

Code from http://www.gamedev.net/topic/597393-getting-the-height-of-a-point-on-a-triangle/ - carefull about computer graphics vs maths use of Y Z

ps. I Don't know of any faster version using shaders. One quick dirty+solution is to render the triangle using colors based on the height of the vertices and pick the pixel color at your X,Y - in practice this never ends up being much faster on a desktop machine, don't know about opengl-es

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