将 3d 4x4 旋转矩阵转换为 2d [英] Convert 3d 4x4 Rotation Matrix into 2d

问题描述

假设我们有一个 4x4 矩阵，索引如下:

Say we have a 4x4 matrix with indices like so:

00 01 02 03
10 11 12 13
20 21 22 23
30 31 32 33

如何将包含在此矩阵中的旋转数据(忽略 z 轴，如果有帮助)转换为单个 2d 旋转角度(以弧度为单位)?

How does one convert the rotation data (ignoring the z axis, if that helps) contained in this matrix into a single 2d rotational angle (in radians)?

背景:我有一个从 Blender 导出为 Collada 格式的 3D .dae 动画.动画在技术上是 2d，所有 z 轴值都是 0.我正在尝试将 4x4 矩阵转换为 2d 平移、旋转和缩放数据.

Background: I have a 3D .dae animation exported from Blender into the Collada format. The animation is technically 2d, all of the z axis values are 0. I'm trying to convert the 4x4 matrices into 2d translation, rotation and scale data.

推荐答案

Scale matrix S 看起来像这样:

Scale matrix S looks like this:

sx 0  0  0
0  sy 0  0
0  0  sz 0
0  0  0  1

翻译矩阵T看起来像这样:

Translation matrix T looks like this:

1  0  0  0
0  1  0  0
0  0  1  0
tx ty tz 1

Z 轴旋转矩阵 R 看起来像这样:

Z-axis rotation matrix Rlooks like this:

 cos(a) sin(a)  0  0
-sin(a) cos(a)  0  0
   0      0     1  0
   0      0     0  1

如果你有一个变换矩阵 M，它是 R、T 和 S 多次相乘的结果 矩阵.查看 M，这些乘法的顺序和数量是未知的.但是，如果我们假设 M=S*R*T 我们可以将其分解为单独的矩阵.首先让我们计算S*R*T:

If you have a transformation matrix M, it is a result of a number of multiplications of R, T and S matrices. Looking at M, the order and number of those multiplications is unknown. However, if we assume that M=S*R*T we can decompose it into separate matrices. Firstly let's calculate S*R*T:

        ( sx*cos(a) sx*sin(a) 0  0)       (m11 m12 m13 m14)
S*R*T = (-sy*sin(a) sy*cos(a) 0  0) = M = (m21 m22 m23 m24)
        (     0         0     sz 0)       (m31 m32 m33 m34)
        (     tx        ty    tz 1)       (m41 m42 m43 m44)

因为我们知道这是一个 2D 转换，所以翻译很简单:

Since we know it's a 2D transformation, getting translation is straightforward:

translation = vector2D(tx, ty) = vector2D(m41, m42)

计算旋转和缩放，我们可以使用sin(a)^2+cos(a)^2=1:

To calculate rotation and scale, we can use sin(a)^2+cos(a)^2=1:

(m11 / sx)^2 + (m12 / sx)^2 = 1
(m21 / sy)^2 + (m22 / sy)^2 = 1

m11^2 + m12^2 = sx^2
m21^2 + m22^2 = sy^2

sx = sqrt(m11^2 + m12^2)
sy = sqrt(m21^2 + m22^2)

scale = vector2D(sx, sy)

rotation_angle = atan2(sx*m22, sy*m12)

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