math - 正交投影矩阵的公式？

Question

我一直环顾四周，似乎无法找到我正在寻找的东西。我找到了“规范公式”，但是使用这些公式的最佳方法是什么？我是否必须缩小每个顶点？ 或者，还有更好的方法？

一个公式真的可以帮助我，但我也在寻找关于相对于观察者位置的近和远 z 平面的解释

Answer 1

这是一个合理的来源，它派生了 orthogonal project matrix :

Consider a few points: First, in eye space, your camera is positioned at the origin and looking directly down the z-axis. And second, you usually want your field of view to extend equally far to the left as it does to the right, and equally far above the z-axis as below. If that is the case, the z-axis passes directly through the center of your view volume, and so you have r = –l and t = –b. In other words, you can forget about r, l, t, and b altogether, and simply define your view volume in terms of a width w, and a height h, along with your other clipping planes f and n. If you make those substitutions into the orthographic projection matrix above, you get this rather simplified version:

以上所有内容都为您提供了一个看起来像这样的矩阵(如果您希望生成的转换矩阵处理任意相机位置和方向，请酌情添加旋转和平移)。


(来源: codeguru.com)