如何确定相机方向与平面的交点?

问题描述

我有一个3D场景，它沿Y垂直轴的高度H为无限大的水平面(与xz坐标平行).

I have a 3D scene with an infinite horizontal plane (parallel to the xz coordinates) at a height H along the Y vertical axis.

我想知道如何确定相机轴线与该平面之间的交点.

I would like to know how to determine the intersection between the axis of my camera and this plane.

摄像机由视图矩阵和投影矩阵定义.

The camera is defined by a view-matrix and a projection-matrix.

推荐答案

这里有两个子问题:1)从摄像机矩阵中提取位置和视图方向. 2)计算视线与平面的交点.

There are two sub-problems here: 1) Extracting the position and view-direction from the camera matrix. 2) Calculating the intersection between the view-ray and the plane.

提取位置和视图方向

视图矩阵描述了点如何从世界空间转换为视图空间.通常在OpenGL中定义视图空间，以使摄影机位于原点并朝-z方向看.

The view matrix describes how points are transformed from world-space to view space. The view-space in OpenGL is usually defined such that the camera is in the origin and looks into the -z direction.

要获取相机的位置，我们必须将视图空间的原点[0,0,0]转换回世界空间.从数学上来说，我们必须计算:

To get the position of the camera, we have to transform the origin [0,0,0] of the view-space back into world-space. Mathematically speaking, we have to calculate:

camera_pos_ws = inverse(view_matrix) * [0,0,0,1]

但是当查看方程式时，我们会发现我们只被插入逆矩阵的第四列，该矩阵将包含

but when looking at the equation we'll see that we are only interrested in the 4th column of the inverse matrix which will contain

camera_pos_ws = [-view_matrix[12], -view_matrix[13], -view_matrix[14]]

可以通过类似的计算找到相机的方向.我们知道相机在视空间中朝-z方向看，因此世界空间方向由

The orientation of the camera can be found by a similar calculation. We know that the camera looks in -z direction in view-space thus the world space direction is given by

camera_dir_ws = inverse(view_matrix) * [0,0,-1,0];

同样，当我们查看方程式时，我们将看到这仅考虑了逆矩阵的第三行，该行由

Again, when looking at the equation, we'll see that this only takes the third row of the inverse matrix into account which is given by

camera_dir_ws = [-view_matrix[2], -view_matrix[6], -view_matrix[10]]

计算交叉点

我们现在知道摄像机的位置P和视角D，因此我们必须沿着光线R(x,y,z) = P + l * D找到x，z值，其中y等于H.由于只有一个未知数l，我们可以计算出来自

We now know the camera position P and the view direction D, thus we have to find the x,z value along the ray R(x,y,z) = P + l * D where y equals H. Since there is only one unknown, l, we can calculate that from

y = Py + l * Dy
H = Py + l * Dy
l = (H - Py) / Dy

然后通过将l重新粘贴到射线方程中来给出交点.

The intersection point is then given by pasting l back into the ray equation.

注释

索引假定矩阵存储在以列为主的线性数组中.

The indices assume that the matrix is stored in a column-major linear array.

注意，形式为矩阵的逆

Note, that the inverse of a matrix of the form

M = [  R T ]
       0 1

，其中R是3x3正交矩阵，由给出

, where R is a orthogonal 3x3 matrix, is given by

inv(M) = [ transpose(R)  -T ]
                0         1

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