使用透视相机矩阵将3D点投影到2D屏幕空间 [英] Projecting a 3D point to 2D screen space using a perspective camera matrix

问题描述

我尝试使用透视相机矩阵将一系列3D点投影到屏幕上。我没有世界空间（或者认为它是一个单位矩阵），我的相机没有相机空间（或认为它是一个单位矩阵），我有一个4x4的矩阵为我的对象空间。

我使用对象矩阵并将其乘以相机透视矩阵，使用以下方法生成：

  Matrix4 createPerspectiveMatrix（Float fov，Float aspect，Float near，Float far）
 {
 Float fov2 =（fov / 2）*（Math.PI / 180） 
 Float tan = Math.tan（fov2）; 
 Float f = 1 / tan; 
 
 return new Matrix4（
f / aspect，0，0，0，
 0，f，0，0，
 0，0，远）/（近远）），（2 *远*近）/（近远），
 0,0,1,0 
）。 
} 
  

然后我将点[x，y，z，1]乘以透视矩阵和对象矩阵的乘积。

下一部分是我困惑的地方，我很确定我需要得到这些点在-1和1或0和1的范围内，并且在具有第一组值的情况下，然后将点乘以屏幕坐标x和y值的屏幕宽度和高度相应地或乘以屏幕高度/ 2和宽度/ 2的值，并将相同的值添加到相应的点。

任何一步一步告诉我这可能是什么实现或在哪里，我可能会错误与任何这将被massivly赞赏！ ：D

最适合所有人！

在身份/翻译矩阵的例子中，我的模型中的矩阵格式是：

  [1,0,0，tx ，
 0，1，0，ty，
 0,0,1，tz，
 0，0，0，1] 
  pre> 
 
解决方案
您的问题是您忘记执行透视划分。
 
 
 
透视除法意味着您将点的x，y和z分量除以其w分量。 
将点从均匀4D空间转换为标准化设备坐标系（NDCS），其中每个分量x，y或z在-1 
 
 
 
此变换后，您可以进行视口转换（按屏幕宽度，高度等乘以点）。
 
 
 
在Foley的书（Computer Graphics：Principles and Practice in C）中有一个良好的转换流程，你可以在这里看到：
 
 
 
  http://www.ugrad.cs.ubc.ca /~cs314/notes/pipeline.html  
 
I am attempting to project a series of 3D points onto the screen using a perspective camera matrix. I do not have world space (or consider it being an identity matrix) and my camera does not have camera space (or consider it an identity matrix), I do have a 4x4 matrix for my object space.

I am taking the object matrix and multiplying it by the camera perspective matrix, generated with the following method:
Matrix4 createPerspectiveMatrix( Float fov, Float aspect, Float near, Float far )
{
    Float fov2 = (fov/2) * (Math.PI/180);
    Float tan = Math.tan(fov2);
    Float f = 1 / tan;

    return new Matrix4 ( 
        f/aspect, 0, 0, 0,
        0, f, 0, 0,
        0, 0, -((near+far)/(near-far)), (2*far*near)/(near-far),
        0, 0, 1, 0 
    );
}
I am then taking my point [x, y, z, 1] and multiplying that by the resulting multiplication of the perspective matrix and object matrix.

The next part is where i'm getting confuzzled, I'm pretty sure that I need to get these points within the ranges of either -1 and 1, or 0 and 1, and, in the case of having the first set of values, I would then multiply the points by the screen width and height for the screen coordinates x and y value respectivly or multiply the values by the screen height/2 and width/2 and add the same values to the respective points.

Any step by step telling me how this might be achieved or where I might be going wrong with any of this would be massivly appreciated!! :D

Best regards everyone!

P.S. In the example of a identity/translation matrix, my matrix format in my model is:
[1, 0, 0, tx,
 0, 1, 0, ty,
 0, 0, 1, tz,
 0, 0, 0, 1 ]

解决方案
Your problem is that you forget to perform the perspective division.

Perspective division means that you divide x, y and z component of your point by its w component.
This is required for transforming your point from Homogeneous 4D Space to Normalized Device Coordinates System (NDCS) in which each component x, y or z falls between -1 and 1, or 0 and 1.

After this transformation, you can do your viewport transformation (multiply points by screen width, heigth etc).

There's a good view of this transformation pipeline in Foley's book (Computer Graphics: Principles and Practice in C), you can see it here:

http://www.ugrad.cs.ubc.ca/~cs314/notes/pipeline.html

                        
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