三维变换矩阵实战——三维点云的旋转、缩放、镜像、错切、平移、正交投影

一、旋转矩阵（右手坐标系）

绕x轴旋转

旋转矩阵：右边矩阵是点云的原始坐标，左边的是旋转矩阵

$\begin{bmatrix} 1&0 &0 \\ 0&cos\beta &-sin\beta \\ 0&sin\beta & cos\beta \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}$

可视化：绕x轴旋转90度

代码：

python">import vtk
import numpy as np
import mathdef pointPolydataCreate(pointCloud):points = vtk.vtkPoints()cells = vtk.vtkCellArray()i = 0for point in pointCloud:points.InsertPoint(i, point[0], point[1], point[2])cells.InsertNextCell(1)cells.InsertCellPoint(i)i += 1PolyData = vtk.vtkPolyData()PolyData.SetPoints(points)PolyData.SetVerts(cells)mapper = vtk.vtkPolyDataMapper()mapper.SetInputData(PolyData)actor = vtk.vtkActor()actor.SetMapper(mapper)actor.GetProperty().SetColor(0.0, 0.1, 1.0)return actordef visiualize(pointCloud, pointCloud2):colors = vtk.vtkNamedColors()actor1 = pointPolydataCreate(pointCloud)actor2 = pointPolydataCreate(pointCloud2)Axes = vtk.vtkAxesActor()# 可视化renderer1 = vtk.vtkRenderer()renderer1.SetViewport(0.0, 0.0, 0.5, 1)renderer1.AddActor(actor1)renderer1.AddActor(Axes)renderer1.SetBackground(colors.GetColor3d('skyblue'))renderer2 = vtk.vtkRenderer()renderer2.SetViewport(0.5, 0.0, 1.0, 1)renderer2.AddActor(actor2)renderer2.AddActor(Axes)renderer2.SetBackground(colors.GetColor3d('skyblue'))renderWindow = vtk.vtkRenderWindow()renderWindow.AddRenderer(renderer1)renderWindow.AddRenderer(renderer2)renderWindow.SetSize(1040, 880)renderWindow.Render()renderWindow.SetWindowName('PointCloud')renderWindowInteractor = vtk.vtkRenderWindowInteractor()renderWindowInteractor.SetRenderWindow(renderWindow)renderWindowInteractor.Initialize()renderWindowInteractor.Start()pointCloud = np.loadtxt("C:/Users/A/Desktop/pointCloudData/model.txt") #读取点云数据angel_x = 90  # 旋转角度
radian = angel_x * np.pi / 180  # 旋转弧度
Rotation_Matrix_1 = [  # 绕x轴三维旋转矩阵 [1, 0, 0],[0, math.cos(radian), -math.sin(radian)],[0, math.sin(radian), math.cos(radian)]]Rotation_Matrix_1 = np.array(Rotation_Matrix_1)p = np.dot(Rotation_Matrix_1, pointCloud.T) # 计算
p = p.T
visiualize(pointCloud, p)

绕y轴旋转

旋转矩阵：

$\begin{bmatrix} cos\beta &0 &sin\beta \\ 0&1 &0 \\ -sin\beta &0 &cos\beta \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}$

可视化：绕y轴旋转180度

代码：

python">angel_y = 180  # 旋转角度
radian = angel_y * np.pi / 180  # 旋转弧度
Rotation_Matrix_2 = [  # 绕y轴三维旋转矩阵[math.cos(radian), 0, math.sin(radian)],[0, 1, 0],[-math.sin(radian), 0, math.cos(radian)]]Rotation_Matrix_1 = np.array(Rotation_Matrix_1)p = np.dot(Rotation_Matrix_1, pointCloud.T) # 计算
p = p.T
visiualize(pointCloud, p)

绕z轴旋转

旋转矩阵：

$\begin{bmatrix} cos\beta &-sin\beta &0 \\ sin\beta &cos\beta & 0\\ 0&0 & 1 \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}$

可视化：绕z轴旋转90度

代码：

python">angel_z = 90  # 旋转角度
radian = angel_z * np.pi / 180  # 旋转弧度
Rotation_Matrix_1 = [  # 绕z轴三维旋转矩阵[math.cos(radian), -math.sin(radian), 0],[math.sin(radian), math.cos(radian), 0],[0, 0, 1]]Rotation_Matrix_1 = np.array(Rotation_Matrix_1)p = np.dot(Rotation_Matrix_1, pointCloud.T) # 计算
p = p.T
visiualize(pointCloud, p)

线绕z轴旋转，再绕x轴旋转：

旋转矩阵: 线绕哪个轴转，xyz矩阵就和哪和轴的旋转矩阵先计算

$\begin{bmatrix} 1&0 &0 \\ 0&cos\beta &-sin\beta \\ 0&sin\beta & cos\beta \end{bmatrix}$ $\begin{bmatrix} cos\beta &-sin\beta &0 \\ sin\beta &cos\beta & 0\\ 0&0 & 1 \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}$

可视化：先绕z轴旋转90度，再绕x轴旋转90度

代码：

python">angel_z = 90  # 旋转角度
radian = angel_z * np.pi / 180  # 旋转弧度
Rotation_Matrix_z = [  # 绕z轴三维旋转矩阵[math.cos(radian), -math.sin(radian), 0],[math.sin(radian), math.cos(radian), 0],[0, 0, 1]]angel_x = 90  # 旋转角度
radian = angel_x * np.pi / 180  # 旋转弧度
Rotation_Matrix_x = [  # 绕x轴三维旋转矩阵[1, 0, 0],[0, math.cos(radian), -math.sin(radian)],[0, math.sin(radian), math.cos(radian)]]Rotation_Matrix_z = np.array(Rotation_Matrix_z)
Rotation_Matrix_x = np.array(Rotation_Matrix_x)p = np.dot(Rotation_Matrix_z, pointCloud.T) # 计算
p = np.dot(Rotation_Matrix_x, p) # 计算
p = p.T
visiualize(pointCloud, p)

二、缩放矩阵

缩放矩阵：

$\begin{bmatrix} k_{x}&0 &0 \\ 0& k_{y} &0 \\ 0&0 &k_{z} \end{bmatrix}$

计算过程：三个k是xyz对应的缩放系数

$\begin{bmatrix} k_{x}&0 &0 \\ 0& k_{y} &0 \\ 0&0 &k_{z} \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}=$ $\begin{bmatrix} k_{x}x\\k_{y} y \\k_{z}z \end{bmatrix}$

x坐标变为原来的1.5倍，y变为0.7倍，z不变

$\begin{bmatrix} 1.5&0 &0 \\ 0& 0.7 &0 \\ 0&0 &1 \end{bmatrix}$ $\begin{bmatrix} x\\ y \\z \end{bmatrix}=$ $\begin{bmatrix} 1.5x\\0.7 y \\1z \end{bmatrix}$

可视化：

三、镜像矩阵

3D镜像矩阵：

$\begin{bmatrix} 1-2n_x{}^{2} & -2n_x{}n_{y} &-2n_x{}n_{z} \\ -2n_x{}n_{y} &1-2n_y{}^{2} &-2n_y{}n_{z} \\ -2n_x{}n_{z} &-2n_y{}n_{z} & 1-2n_z{}^{2} \end{bmatrix}$

$n=\begin{bmatrix} n_{x} &n_{y} & n_{z} \end{bmatrix}$

向量n是垂直于镜像平面的单位向量

三维点云对xz平面的镜像：

①首先，确定一个垂直于xz平面的单位向量 n=[0, 1, 0]

②将该单位向量带入上述3D镜像矩阵

可视化：

代码：

python">import vtk
import numpy as np
import mathdef pointPolydataCreate(pointCloud):points = vtk.vtkPoints()cells = vtk.vtkCellArray()i = 0for point in pointCloud:points.InsertPoint(i, point[0], point[1], point[2])cells.InsertNextCell(1)cells.InsertCellPoint(i)i += 1PolyData = vtk.vtkPolyData()PolyData.SetPoints(points)PolyData.SetVerts(cells)mapper = vtk.vtkPolyDataMapper()mapper.SetInputData(PolyData)actor = vtk.vtkActor()actor.SetMapper(mapper)actor.GetProperty().SetColor(0.0, 0.1, 1.0)return actordef visiualize(pointCloud, pointCloud2):colors = vtk.vtkNamedColors()actor1 = pointPolydataCreate(pointCloud)actor2 = pointPolydataCreate(pointCloud2)Axes = vtk.vtkAxesActor()# 可视化renderer1 = vtk.vtkRenderer()renderer1.SetViewport(0.0, 0.0, 0.5, 1)renderer1.AddActor(actor1)renderer1.AddActor(Axes)renderer1.SetBackground(colors.GetColor3d('skyblue'))renderer2 = vtk.vtkRenderer()renderer2.SetViewport(0.5, 0.0, 1.0, 1)renderer2.AddActor(actor1)renderer2.AddActor(actor2)renderer2.AddActor(Axes)renderer2.SetBackground(colors.GetColor3d('skyblue'))renderWindow = vtk.vtkRenderWindow()renderWindow.AddRenderer(renderer1)renderWindow.AddRenderer(renderer2)renderWindow.SetSize(1040, 880)renderWindow.Render()renderWindow.SetWindowName('PointCloud')renderWindowInteractor = vtk.vtkRenderWindowInteractor()renderWindowInteractor.SetRenderWindow(renderWindow)renderWindowInteractor.Initialize()renderWindowInteractor.Start()pointCloud = np.loadtxt("C:/Users/A/Desktop/pointCloudData/model.txt") #读取点云数据nx = 0
ny = 0
nz = 1
n = [nx, ny, nz] # 垂直xy平面的单位向量
# 镜像矩阵
Mirror_Matrix = [[1-2*nx**2, -2*nx*ny, -2*nx*nz],[-2*nx*ny, 1-2*ny**2, -2*ny*nz],[-2*nx*nz, -2*ny*nz, 1-2*nz**2]]Mirror_Matrix = np.array(Mirror_Matrix)p = np.dot(Mirror_Matrix, pointCloud.T)  # 计算
p = p.T
visiualize(pointCloud, p)

四、错切矩阵

沿xy平面错切（z不变）

矩阵计算过程

$H_{xy}(s, t)=\begin{bmatrix} 1 &0 &s \\ 0&1 &t \\ 0&0 &1 \end{bmatrix}$ $\begin{bmatrix} 1 &0 &s \\ 0&1 &t \\ 0&0 &1 \end{bmatrix} \begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} x+sz\\y+tz \\ z \end{bmatrix}$

沿xz平面错切（y不变）

矩阵计算过程

$H_{xz}(s, t)=\begin{bmatrix} 1 &0 &s \\ 0&1 &0 \\ 0&t &1 \end{bmatrix}$ $\begin{bmatrix} 1 &0 &s \\ 0&1 &0 \\ 0&t &1 \end{bmatrix} \begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} x+sz\\y \\ z+ty \end{bmatrix}$

沿yz平面错切（x不变）

矩阵计算过程

$H_{yz}(s, t)=\begin{bmatrix} 1 &0 &0 \\ s&1 &0 \\ t&0 &1 \end{bmatrix}$ $\begin{bmatrix} 1 &0 &0 \\ s&1 &0 \\ t&0 &1 \end{bmatrix} \begin{bmatrix} x\\ y \\ z \end{bmatrix}=\begin{bmatrix} x\\y+sx \\ z+tx \end{bmatrix}$

可视化：沿yz平面错切

代码：

python">pointCloud = np.loadtxt("C:/Users/A/Desktop/pointCloudData/model.txt") #读取点云数据s = 0.3
t = 0.3
# 沿yz平面错切矩阵
Shear_Matrix = [[1, 0, 0],[s, 1, 0],[t, 0, 1]]Shear_Matrix = np.array(Shear_Matrix)p = np.dot(Shear_Matrix, pointCloud.T)  # 计算
p = p.T
visiualize(pointCloud, p)

五、正交投影

正交投影矩阵（投影到三维空间任意平面）：

$\begin{bmatrix} 1-n_{x}^{2} &-n_{x}n_{y} &-n_{x}n_{z} \\ -n_{x}n_{y}&1-n_{y}^{2} &-n_{y}n_{z} \\ -n_{x}n_{z}&-n_{y}n_{z} & 1-n_{z}^{2} \end{bmatrix}$

$n=\begin{bmatrix} n_{x} &n_{y} & n_{z} \end{bmatrix}$

向量n是垂直于投影平面的单位向量

可视化：点云在xy平面上的正交投影

六、平移矩阵

平移矩阵需要利用齐次矩阵(4*4矩阵)，下面是一个平移矩阵

最右边一列是xyz的位移量

$\begin{bmatrix} 1 &0 &0 & \Delta x \\ 0&1 &0 &\Delta y \\ 0&0 &1 & \Delta z\\ 0&0 & 0 & 1 \end{bmatrix}$

计算过程：

$\begin{bmatrix} 1 &0 &0 & \Delta x \\ 0&1 &0 &\Delta y \\ 0&0 &1 & \Delta z\\ 0&0 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} x\\y \\z \\ 1 \end{bmatrix}=\begin{bmatrix} x+\Delta x\\ y+\Delta y \\ z+\Delta z \\ 1 \end{bmatrix}$