透视变换矩阵的计算 [英] Calculation of a perspective transformation matrix

问题描述

由于在三维空间中的一个点，我如何计算在齐次坐标矩阵，将预计，点到面以Z = D ，这里的原点是中心投影。

Given a point in 3D space, how can I calculate a matrix in homogeneous coordinates which will project that point into the plane z == d, where the origin is the centre of projection.

推荐答案

OK，让我们试着整理了这一点，不断扩大对伊曼纽尔的答案。

OK, let's try to sort this out, expanding on Emmanuel's answer.

假设的，你的观点载体直接沿Z轴，所有维度都必须由视图平面的距离 D 来的比例缩放原以Z 坐标。这比平凡 D / Z ，赠送：

Assuming that your view vector is directly along the Z axis, all dimensions must be scaled by the ratio of the view plane distance d to the original z coordinate. That ratio is trivially d / z, giving:

x' = x * (d / z)
y' = y * (d / z)
z' = z * (d / z)    ( = d)

在齐次坐标，它通常以启动与P = [X，Y，Z，W] ，其中W¯¯== 1 和转型因此完成的：

In homogenous coordinates, it's usual to start with P = [x, y, z, w] where w == 1 and the transformation is done thus:

P' = M * P

结果将有W¯¯！= 1 ，并获得真正的三维坐标，我们标准化的同质载体通过将整个事情由它的是W 部分。

The result will have w != 1, and to get the real 3D coordinates we normalise the homogenous vector by dividing the whole thing by its w component.

所以，我们需要的是给定一个矩阵 [X，Y，Z，1 ]为我们提供了 [X * D，Y * D，Z * D，Z] ，即

So, all we need is a matrix that given [x, y, z, 1] gives us [x * d, y * d, z * d, z], i.e.

| x' |  =    | d   0   0   0 |  *  | x |
| y' |  =    | 0   d   0   0 |  *  | y |
| z' |  =    | 0   0   d   0 |  *  | z |
| w' |  =    | 0   0   1   0 |  *  | 1 |

曾经归（由 W'分裂==ž）为您提供了：

[ x * d / z, y * d / z,   d,   1 ]

每第一组方程式以上

per the first set of equations above

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