Finding the linear mapping between homogeneous coordinates

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The discussion focuses on deriving a linear mapping between world coordinates (X, Y, Z) and image coordinates (x, y) in the context of an affine camera model. The governing equation involves a 2x3 matrix A and a 2x1 vector b, which represent the projection relationship. Participants clarify that while perspective projection uses a 3x4 matrix, the affine projection requires a different approach due to the absence of depth information. The conversation emphasizes starting with an identity matrix for the first three columns and considering the impact of the fourth column's non-zero value. Understanding these components is crucial for accurately forming the desired mapping.
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Homework Statement


If I have an affine camera with a projection relationship governed by:

\begin{equation}
\begin{bmatrix}
x & y
\end{bmatrix}^T = A
\begin{bmatrix}
X & Y & Z
\end{bmatrix}^T + b
\end{equation}
where A is a 2x3 matrix and b is a 2x1 vector. How can I form a matrix representing the linear mapping between the world point (X,Y,Z) and image point (x,y) if they are represented by homogeneous vectors?

Homework Equations


NIL

The Attempt at a Solution


I understand that for a camera perspective projection, the linear mapping between homogeneous coordinates where the equation is only up to a scale factor can be written as a 3x4 projection matrix that represents a map from 3D to 2D.
\begin{equation}
\begin{bmatrix}
x \\
y \\
w
\end{bmatrix}=
\begin{bmatrix}
f & 0 & 0 & 0 \\
0 & f & 0 & 0 \\
0 & 0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
X_c \\
Y_c \\
Z_c \\
1
\end{bmatrix}
\end{equation}
However, I am unclear as to how an affine camera differs and how to fit a linear mapping based on the above given governing equation.
 
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Remember you want your original vector plus the displacement. Start with the identity matrix components in the first three columns and then ask what effect a non-zero value in the fourth column will have.
 

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