将 OBJ 文件中的四边形转换为三角形? [英] Converting quadriladerals in an OBJ file into triangles?

问题描述

起初，这似乎很明显......在找到 4 个索引的地方为每个面制作 2 个三角形，对吗?
含义如下:

At first, it seemed obvious... Make 2 triangles per face wherever 4 indices were found, right?
Meaning, the following:

v 1.000000 1.000000 0.000000
v -1.000000 1.000000 -0.000000
v 1.000000 -1.000000 0.000000
v -1.000000 -1.000000 -0.000000
f -4 -3 -2 -1

... 反过来，需要转换成类似的东西:

... would, in turn, need to be converted into something like:

v 1.000000 1.000000 0.000000
v -1.000000 1.000000 -0.000000
v 1.000000 -1.000000 0.000000
v -1.000000 -1.000000 -0.000000
f -4 -3 -2
f -2 -3 -1

当然，这个特殊的例子会正确呈现.
但是，并非所有情况都像将人脸分成两个人脸一样简单(其中第一个人脸包含原始人脸的前三个顶点，第二个人脸包含最后三个顶点，如上例所示).以下面的立方体为例:

This particular example, of course, would render correctly.
However, not all cases are as simple as splitting the face into two faces (where the first face contains the first three vertices of the original face, and the second face contains the last 3 vertices, as per the above example). Take the following cube, for example:

v 0.000000 1.000000 1.000000
v 0.000000 0.000000 1.000000
v 1.000000 0.000000 1.000000
v 1.000000 1.000000 1.000000
v 0.000000 1.000000 0.000000
v 0.000000 0.000000 0.000000
v 1.000000 0.000000 0.000000
v 1.000000 1.000000 0.000000
f 1 2 3 4
f 8 7 6 5
f 4 3 7 8
f 5 1 4 8
f 5 6 2 1
f 2 6 7 3

这些面不能在前面的例子中以同样的方式分割......所以，我需要一些方法来知道如何将一个四边形面分成两个三角形面，同时使用正确的索引对于第二张脸...

These faces cannot be split the same way in the previous example... So, I would need some way of knowing how to split a quadrilateral face into two triangle faces, whilst using the correct indices for the second face...

如何实现?请注意，我没有使用固定功能管道，因此，使用 GL_QUADS 不是一种选择.我的渲染引擎几乎只能使用 GL_TRIANGLES.

How can this be achieved? Please note that I am NOT using the fixed-function pipeline, and therefore, using GL_QUADS is NOT an option. My rendering engine is pretty much stuck on using GL_TRIANGLES only.

推荐答案

如果你有 4 个索引，例如:

If you have 4 indices, e.g.:

0 1 2 3

划分为两个三角形将是一个具有前 3 个索引，一个具有第一个、第三个和第四个.在这个例子中:

The division into two triangles would be one with the first 3 indices, and one with the first, third, and fourth. In this example:

0 1 2
0 2 3

让我们尝试一些 ASCII 艺术来说明这一点:

Let's try some ASCII art to illustrate this:

3-------2
|      /|
|    /  |
|  /    |
|/      |
0-------1

这里你看到 0 1 2 3 是四边形，0 1 2 是第一个三角形(右下角)，0 2 3代码>作为第二个三角形(左上角).

Here you see 0 1 2 3 as the quad, 0 1 2 as the first triangle (bottom-right), and 0 2 3 as the second triangle (top left).

更一般地，对于具有 n 个顶点的面，您会生成三角形:

More generally, for faces with n vertices, you generate triangles:

0 (i) (i + 1)  [for i in 1..(n - 2)]

如果你不坚持单独的三角形，你也可以使用 GL_TRIANGLE_FAN 基元，它们仍然在核心 OpenGL 中.这样，您可以使用原始索引序列绘制带有三角扇形的任何凸多边形.因此，顶点序列 0 1 2 3 的三角形扇形描述了这种情况下的四边形，并且很容易推广到具有 4 个以上顶点的面.

If you don't insist on separate triangles, you can also use GL_TRIANGLE_FAN primitives, which are still in core OpenGL. That way, you can draw any convex polygon with a triangle fan, using the original sequence of indices. So a triangle fan with vertex sequence 0 1 2 3 describes the quad in this case, and it very easily generalizes to faces with more than 4 vertices.

由于您似乎仍有问题，让我们看看这如何适用于您帖子中的示例.我将列出每个面的四边形的原始索引序列，以及分裂四边形后两个三角形的索引序列.

Since you still appear to have problems, let's see how this applies to the example in your post. I'll list the original index sequence of the quad for each face, and the index sequence for the two triangles after splitting the quad.

f 1 2 3 4 --> (1 2 3) (1 3 4)
f 8 7 6 5 --> (8 7 6) (8 6 5)
f 4 3 7 8 --> (4 3 7) (4 7 8)
f 5 1 4 8 --> (5 1 4) (5 4 8)
f 5 6 2 1 --> (5 6 2) (5 2 1)
f 2 6 7 3 --> (2 6 7) (2 7 3)

当我绘制立方体时，这对我来说是正确的.请记住从索引中减去 1 以供您使用，因为这些是从 1 开始的索引，您几乎肯定需要从 0 开始的索引.

That looks correct to me when I draw the cube. Remember to subtract 1 from the indices for your use, since these are 1-based indices, and you will almost certainly need 0-based indices.

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