3D平面点云与2D坐标双向转换：求变换矩阵T1、T2

Alright, let's work through this problem step by step. Since all your points lie on a single 3D plane, we can leverage planar geometry to define a custom 2D coordinate system (B) and derive the transformation matrices T1 (3D A → 2D B) and T2 (2D B → 3D A) easily.

Step 1: Characterize the Plane in Coordinate System A #

First, we need to define the plane that contains all your point cloud points in system A. You can do this by:

• Calculating the plane's normal vector n = (a, b, c): Use PCA on your point cloud—this is the eigenvector corresponding to the smallest eigenvalue of the point cloud's covariance matrix (perfect for handling noisy point data).
• Choosing a origin point P₀ = (x₀, y₀, z₀) on the plane: This can be the centroid of your point cloud, or any arbitrary point sampled directly from the cloud.

The plane's equation is then:

n · (P - P₀) = 0  →  a(x - x₀) + b(y - y₀) + c(z - z₀) = 0

Step 2: Define Custom 2D Coordinate System B #

Since system B can be arbitrary, we'll set it up for simplicity and geometric clarity:

• Origin of B: Map this directly to P₀ (the plane origin we chose in Step 1).
• Basis vectors of B: We need two orthogonal unit vectors lying within the plane to serve as the x'-axis and y'-axis of B:
  1. Pick any vector v that's not parallel to n (e.g., if n ≠ (1,0,0), use v = (1,0,0)).
  2. Compute the first basis vector: u₁ = normalize(v × n) (cross product ensures it's perpendicular to n, hence lies in the plane; normalize to make it unit length).
  3. Compute the second basis vector: u₂ = normalize(n × u₁) (cross product ensures it's perpendicular to both n and u₁, so it's orthogonal to u₁ and lies in the plane).

Now B is fully defined: origin at P₀, x'-axis along u₁, y'-axis along u₂.

Step 3: Derive Transformation Matrix T1 (A → B) #

T1 converts a 3D point P = (x, y, z) in system A to a 2D point P' = (x', y') in system B.

Linear Form (Non-Homogeneous) #

First, express P relative to B's origin: vec = P - P₀.
Then project this vector onto B's basis vectors to get the 2D coordinates:

x' = vec · u₁ = (x - x₀)u₁ₓ + (y - y₀)u₁ᵧ + (z - z₀)u₁_z
y' = vec · u₂ = (x - x₀)u₂ₓ + (y - y₀)u₂ᵧ + (z - z₀)u₂_z

Matrix Form (Homogeneous Coordinates) #

To write this as a single matrix multiplication, use homogeneous coordinates (add a 1 to the 3D point):
T1 is a 2×4 matrix:

T1 = [
  [u₁ₓ, u₁ᵧ, u₁_z, -u₁ · P₀],
  [u₂ₓ, u₂ᵧ, u₂_z, -u₂ · P₀]
]

Applying it:

[x'; y'] = T1 × [x; y; z; 1]

Step 4: Derive Transformation Matrix T2 (B → A) #

T2 converts a 2D point P' = (x', y') in system B back to a 3D point P = (x, y, z) in system A.

Linear Form (Non-Homogeneous) #

We just reverse the projection: the 3D point is B's origin plus the linear combination of B's basis vectors scaled by the 2D coordinates:

P = P₀ + x'*u₁ + y'*u₂

Expanded:

x = x₀ + x'*u₁ₓ + y'*u₂ₓ
y = y₀ + x'*u₁ᵧ + y'*u₂ᵧ
z = z₀ + x'*u₁_z + y'*u₂_z

Matrix Form (Homogeneous Coordinates) #

Use homogeneous coordinates for the 2D point (add a 1):
T2 is a 3×3 matrix:

T2 = [
  [u₁ₓ, u₂ₓ, x₀],
  [u₁ᵧ, u₂ᵧ, y₀],
  [u₁_z, u₂_z, z₀]
]

Applying it:

[x; y; z] = T2 × [x'; y'; 1]

Example for Intuition #

Suppose your plane is the XY-plane (z=0) in system A:

• n = (0,0,1), P₀ = (0,0,0)
• u₁ = (1,0,0), u₂ = (0,1,0)

Then:

• T1 becomes [[1,0,0,0],[0,1,0,0]] (simply drops the z-coordinate, which makes intuitive sense)
• T2 becomes [[1,0,0],[0,1,0],[0,0,0]] (adds a z=0 component to the 2D point)

This matches our intuitive expectation for projecting XY-plane points to 2D.

内容的提问来源于stack exchange，提问作者Ujjwal