沿y轴查看时OpenGL不显示任何内容 [英] OpenGL is not displaying anything when looking along the y axis

问题描述

我目前正在尝试使用SFML和OpenGL显示从顶部观察的红色网格. 网格平面垂直于y轴(y轴是从顶部看的向量).

I am currently trying to display a red grid viewed from the top using SFML and OpenGL. The grid plan is normal to the y axis (y axis is the vector looking to the top).

这在我的渲染循环之前被调用:

This is called before my rendering loop:

glEnable(GL_DEPTH_TEST);
gluPerspective(90.0f, (GLfloat)640/(GLfloat)480, 1.0f, 1000.0f);

在我的渲染循环中:

glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(0.0, -3.0, 0.0,
    0.0, -0.8, 0.0,
    0.0, 1.0, 0.0);

for (float x = -10.0f, x < 10.0f, x += 1.0f) {
    glColor3f(1.0f, 0.0f, 0.0f);
    glBegin(GL_LINES);
        glVertex3f(x, -0.5f, -10.0f);
        glVertex3f(x, -0.5f, 10.0f);
        glVertex3f(-10.0f, -0.5f, x);
        glVertex3f(10.0f, -0.5f, x);
    glEnd();
}

window.display(); // swap buffers in SFML

我剥去了SFML部分.

I stripped the SFML part.

当我替换gluLookAt()时；一部分(相机略微倾斜):

When I replace the gluLookAt(); part by this one (slight tilt of the camera):

gluLookAt(0.0, -3.0, 0.0,
    0.0, -0.8, 0.01,
    0.0, 1.0, 0.0);

或这一个: gluLookAt(0.0，-3.0，0.0， 0.0，-0.8、0.0， 0.0、0.0、1.0)；

or this one: gluLookAt(0.0, -3.0, 0.0, 0.0, -0.8, 0.0, 0.0, 0.0, 1.0);

一切显示正常.

我得出的结论是，当我精确地沿-y方向(90°)看时，什么也没画.但为什么?我错过了什么吗?

I concluded that when I look exactly in the -y direction (90°), nothing is drawn. But why? Did I miss something?

我的OpenGL支持的版本是2.1 Mesa 10.2.4

My OpenGL supported version is 2.1 Mesa 10.2.4

非常感谢！

推荐答案

问题是，您的上轴也指向y轴.向上矢量不必完全指向上方，但必须与您的观察方向不同.有关更多详细信息，请此答案

The problem is, that you up-axis also points towards the y-axis. It is not necessary that the up-vector points exactly upwards, but it has to be different from your viewing-direction. For more details have a look at this answer

根据要求，以下是数学解释:

Since requested, here is the mathematical explaination:

gluLookAt生成一个3d坐标框架，该框架由三个垂直向量组成. 假设gluLookAt的输入是摄影机位置(C)，眼睛位置(E)和上矢量(U)，然后使用以下公式构造矩阵:

gluLookAt generates a 3d coordinate frame, that consists of three perpendicular vectors. Assuming the input to gluLookAt is a camera position (C), a eye-position (E), and a up-vector (U), than the matrix is constructed using the following equations:

F = C - E
f = F / |F|

U' = U / |U|

这将导致两个向量跨越一个平面.我们已经知道，f(视图矢量)将是我们最终坐标系的-z轴(opengl始终沿负z轴显示).可以通过找到由f和u跨越的平面的法线向量来计算x轴，因此可以通过以下方式进行计算:

this results in two vectors that span a plane. We already know, that f (the view-vector) will be the -z axis of our final coordinate frame (opengl looks always along the negative z-axis). The x-axis can be calculated by finding the normal vector of the plane spanned by f and u and can thus be calculated via

S = f x U' (x is the cross product)
s = S / |S|

请注意，如果u和f已经垂直，则不需要进行归一化，但是由于gluLookAt不需要实向上向量，因此不必这样做.出于同样的原因，我们不能直接将u用作y轴.我们正在寻找的是一个垂直于f和s的y向量，其给出为:

Note that the normalization would not be required if u and f are already perpendicular, but since gluLookAt does not require the real up vector this is not necessarily given. Out of the same reason we cannot use u directly as y-axis. What we are looking for is a y-vector that is perpendicular to both, f and s, which is given as:

u = s x f

最终矩阵是由

    |     s    0 |
M = |     u    0 |
    |    -f    0 |
    | 0  0  0  1 |

现在考虑您的情况，其中f = U': 计算S时得到

Now consider your case where f = U': When calculating S we get

 S = f x U' = U' x U' = [0, 0, 0]

由于这给出了一个零向量，因此u也将是一个零向量.总而言之，这将产生一个矩阵，其中前两行全为零，因此，每个与该矩阵相乘的向量都将以x和y等于0结束.

Since this gives a null-vector, u will also be a null-vector. In total this results in a matrix where the first two rows are all zero, and thus each vector multiplied with this matrix will end up with x and y beeing 0.

希望这对您有帮助. gluLookAt的公式取自此处

Hope this helps you. The formulas for gluLookAt are taken from here

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