如何从投影正方形估算立方体角点？求(0,0,1)投影坐标

Alright, let's break down how to derive the projected image coordinates for the cube corner (0,0,1) using the given projections of the z=0 unit square.

Step 1: Understand the Homogeneous Projection Matrix #

We're working with a 3×4 homogeneous transformation matrix that maps 3D points (in homogeneous coordinates (x,y,z,1)) to 2D image points (in homogeneous coordinates (u',v',w')):

[u']   [a b c d] [x]
[v'] = [e f g h] [y]
[w']   [i j k 1] [z]
                 [1]

The actual image coordinates (u,v) are calculated as (u'/w', v'/w')—we divide by the third component to convert from homogeneous to Cartesian coordinates.

Step 2: Derive Known Matrix Elements from the Z=0 Square Points #

Using the four given projections of the unit square lying on the z=0 plane, we can solve for some of the matrix elements directly:

• For the origin point (0,0,0) which projects to (u₀₀₀, v₀₀₀):
  Substituting into the matrix gives u' = d, v' = h, w' = 1. Since u₀₀₀ = d/1 and v₀₀₀ = h/1, we immediately get:
  d = u₀₀₀ and h = v₀₀₀.
• For the point (1,0,0) which projects to (u₁₀₀, v₁₀₀):
  We have u' = a + d and w' = i + 1. Rearranging the projection equation u₁₀₀ = (a + u₀₀₀)/(i+1) gives:
  a = u₁₀₀*(i+1) - u₀₀₀
  Applying the same logic to the v-coordinate gives:
  e = v₁₀₀*(i+1) - v₀₀₀
• For the point (0,1,0) which projects to (u₀₁₀, v₀₁₀):
  Following the same steps as above:
  b = u₀₁₀*(j+1) - u₀₀₀
f = v₀₁₀*(j+1) - v₀₀₀

Step 3: Solve for i and j Using the (1,1,0) Point #

The fourth square point (1,1,0) projects to (u₁₁₀, v₁₁₀). Substitute the derived expressions for a, b, and d into the u-coordinate projection equation:

u₁₁₀ = (a + b + d) / (i + j + 1)

Replacing a, b, d and simplifying the right-hand side gives us a linear equation in terms of i and j:

i*(u₁₁₀ - u₁₀₀) + j*(u₁₁₀ - u₀₁₀) = u₁₀₀ + u₀₁₀ - u₀₀₀ - u₁₁₀  --- (1)

Repeating this process for the v-coordinate yields a second linear equation:

i*(v₁₁₀ - v₁₀₀) + j*(v₁₁₀ - v₀₁₀) = v₁₀₀ + v₀₁₀ - v₀₀₀ - v₁₁₀  --- (2)

You can solve this system of two linear equations to find i and j entirely using the eight given projection parameters (u₀₀₀, v₀₀₀, u₁₀₀, v₁₀₀, u₀₁₀, v₀₁₀, u₁₁₀, v₁₁₀).

Step 4: Express the (0,0,1) Projection Coordinates #

Now, let's compute the projection for the cube corner (0,0,1). Substitute this point into the transformation matrix:

u' = c*1 + d = c + u₀₀₀
v' = g*1 + h = g + v₀₀₀
w' = k*1 + 1 = k + 1

The projected image coordinates are:

u₀₀₁ = (c + u₀₀₀) / (k + 1)
v₀₀₁ = (g + v₀₀₀) / (k + 1)

As noted in the problem statement, we cannot uniquely solve for c, g, or k without additional data about points along the Z-axis (like another cube corner's projection). However, we can rewrite these equations using relative depth parameters to make them more usable:
Let r = 1/(k+1) (a scaling factor related to the depth of the Z=1 plane) and s = c*r, t = g*r. Then:

u₀₀₁ = s + u₀₀₀*r
v₀₀₁ = t + v₀₀₀*r

If you have extra constraints (e.g., knowing the cube is axis-aligned with unit depth), you could further refine these equations. But with only the Z=0 square projections, s, t, and r remain free parameters that depend on the camera's position and orientation relative to the cube.

内容的提问来源于stack exchange，提问作者Darin