3D 广告牌精灵背后的数学原理是什么?(原为:3D 变换矩阵到 2D 矩阵) [英] What are the maths behind 3D billboard sprites? (was: 3D transformation matrix to 2D matrix)

问题描述

我在空间中有一个 3D 点.点的确切方向/位置通过 4x4 变换矩阵表示.

I have a 3D point in space. The point's exact orientation/position is expressed through a 4x4 transformation matrix.

我想画一个广告牌(3D Sprite)这点.我知道该点的投影位置(即 3D->2D)；广告牌正对着相机，所以这也很有帮助.我不知道广告牌应该具有的缩放比例！

I want to draw a billboard (3D Sprite) to this point. I know the projected position (i.e. 3D->2D) of the point; the billboard is facing the camera so that's very helpful too. What I don't know is the scaling that the billboard should have!

为了让事情变得更复杂，4x4 矩阵可能有各种变换:3D 旋转、3D 缩放、3D 转置.假设相机尽可能简单:定位在 (0,0,0)，不旋转.

To make things more complex, the 4x4 matrix may have all sorts of transformations: 3D rotation, 3D scaling, 3D transposition. Assume that the camera is as simple as it can be: position at (0,0,0), no rotation.

那么，我可以从这个 4x4 矩阵中提取"广告牌精灵的缩放比例吗?

So, can I "extract" the scaling of the billboard sprite from this 4x4 matrix?

我有一个 3D 仿射变换 4x4 矩阵.我需要将它(项目)转换为 2D 仿射变换 3x3 矩阵，如下所示:

I have a 3D affine transformation 4x4 matrix. I need to convert it (project) to a 2D affine transformation 3x3 matrix, which looks like this:

3D 旋转是无关紧要的，如果存在可能会被丢弃；我只对翻译感兴趣，最重要的是缩放.

3D rotations are irrelevant and if present may be discarded; I am only interested in translation and most importantly scaling.

谁能帮忙算出6 4 个值中每一个的方程式?(假设 t_x, t_y 也是已知的)

Can anyone help with the equations for each of the six 4 values? (lets say t_x, t_y are also known)

附加信息:

Matrix3D 是 3D 点的全局变换，例如 (0,0,0).其目的是投影到二维平面(计算机屏幕)上.

The Matrix3D is the global transformation of a 3D point, say (0,0,0). Its purpose is to be projected on a 2D plane (the computer screen).

我知道如何将 3D 点投影到 2D 空间，我正在寻找的是保留位置以外的附加变换信息，即 缩放:如您所知，缩放 在 2D 平面上投影点时，属性也会改变.

I know how to project a 3D point to 2D space, what I am looking for is to preserve additional transformation information beyond position, i.e. scaling: as you may know, the scaling property is also altered when projecting the point on a 2D plane.

我还忘了提到透视投影属性也是已知的，即:

I also forgot to mention that the perspective projection properties are also known, i.e.:

field of view (single value)
focal length (single value)
projection center (viewpoint position - 2D value)

推荐答案

如果你不使用球坐标系，那么这个任务是不可解决的，因为在投影前丢弃 Z 坐标会移除与投影点的距离，因此你不会知道如何应用透视.

if you not using spherical coordinate system then this task is not solvable because discarding Z-coordinate before projection will remove the distance form the projection point and therefore you do not know how to apply perspective.

你有两个选择(除非我忽略了什么):

You have two choices (unless I overlooked something):

1. 应用 3D 变换矩阵

然后只使用 x,y - 结果的坐标

and then use only x,y - coordinates of the result

为旋转/投影创建 3x3 变换矩阵

并在应用它之前或之后添加偏移向量.请注意，这种方法不使用齐次坐标！！！

and add offset vector before or after applying it. Be aware that this approach do not use homogenous coordinates !!!

等式以求清晰

不要忘记 3x3 矩阵 + 向量变换不是累积的！！！这就是使用 4x4 转换的原因.现在你可以扔掉最后一行矩阵/向量 (Xz,Yz,Zz), (z0) 然后输出向量就是 (x', y').当然在这之后你不能使用逆变换，因为你失去了 Z 坐标.

Do not forget that 3x3 matrix + vector transforms are not cumulative !!! That is the reason why 4x4 transforms are used instead. Now you can throw away the last row of matrix/vector (Xz,Yz,Zz), (z0) and then the output vector is just (x', y'). Of course after this you cannot use the inverse transform because you lost Z coordinate.

缩放是通过改变轴方向向量的大小来完成的

顺便说一句.如果您的投影平面也是 XY-没有旋转的平面，则:

Btw. if your projection plane is also XY-plane without rotations then:

x' = (x-x0)*d/(z-z0)
y' = (y-y0)*d/(z-z0)

(x,y,z) - 指向项目
(x',y') - 投影点
(x0,y0,z0) - 投影原点
d - 焦距

(x,y,z) - point to project
(x',y') - projected point
(x0,y0,z0) - projection origin
d - focal length

[Edit2] 问题编辑后意义完全不同

我假设您希望精灵始终面向相机.它很丑，但简化了草，树，...

I assume you want sprite always facing camera. It is ugly but simplifies things like grass,trees,...

M - 你的矩阵
P - M 内的投影矩阵
如果你的 M = (0,0,0) 原点没有旋转/缩放/倾斜，那么 M=P
pnt - 你的广告牌的点(我假设为中心)(w=1)[GCS]
dx,dy - 一半大小的广告牌 [LCS]
A,B,C,D - 广告牌的投影边缘 [GCS]
[GCS] - 全局坐标系
[LCS] - 局部坐标系

M - your matrix
P - projection matrix inside M
If you have origin of M = (0,0,0) without rotations/scaling/skew then M=P
pnt - point of your billboard (center I assume) (w=1) [GCS]
dx,dy - half sizes of billboard [LCS]
A,B,C,D - projected edges of your billboard [GCS]
[GCS] - global coordinate system
[LCS] - local coordinate system

1. 如果你知道投影矩阵

我假设它是 glFrustrum 或 gluPerspective ...然后:

I assume it is glFrustrum or gluPerspective ... then:

(x,y,z,w)=(M*(P^-1))*pnt  // transformed center of billboard without projection
A=P*(x-dx,y-dy,z,w)
B=P*(x-dx,y+dy,z,w)
C=P*(x+dx,y+dy,z,w)
D=P*(x+dx,y-dy,z,w)

如果您的 M 矩阵太复杂，#1 无法工作

If your M matrix is too complex for #1 to work

MM=(M*(P^-1))     // transform matrix without projection
XX=MM(Xx,Xy,Xz)   // X - axis vector from MM [GCS](look at the image above on the right for positions inside matrix)
YY=MM(Yx,Yy,Yz)   // Y - axis vector from MM [GCS]
X =(M^-1)*XX*dx   // X - axis vector from MM [LCS] scaled to dx
Y =(M^-1)*YY*dy   // Y - axis vector from MM [LCS] scaled to dy
A = M*(pnt-X-Y)
B = M*(pnt-X+Y)
C = M*(pnt+X+Y)
D = M*(pnt+X-Y)

[Edit3] 仅缩放

MM=(M*(P^-1))      // transform matrix without projection
sx=|MM(Xx,Xy,Xz)|  // size of X - axis vector from MM [GCS] = scale x
sy=|MM(Yx,Yy,Yz)|  // size of Y - axis vector from MM [GCS] = scale y

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