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Next: Summary of Plane Up: No Title Previous: Example 2: Synthetic

Computing Plane Projective Transformations

Estimate for ,

where = is equality up to scale. Minimum number of correspondences: 4 points (non-collinear) or 4 lines (non-concurrent).


n = 4 will be covered as a special case. Each point correspondence generates two linear equations for the elements of (dividing by the third component to remove the unknown scale factor)

and multiplying out

Then points generates 2 n linear equations, which are sufficient to solve for . The problem with projective transformations is that the matrix is only recovered up to scale.

Homogeneous Solutions There are two methods of dealing with the unknown scale factor in a homogeneous matrix

  1. Choose one of the matrix elements to have a certain value. For example, .
  2. Solve for the matrix up to scale.

Method I

If , then the above equations can be written:

for 4 points. A linear solution is then obtained, in the usual manner, by solving the set of linear simultaneous equations. Similarly, for n > 4 points, a solution can be obtained using a pseudo-inverse.

The problem with this approach is that if the true solution is , then this cannot be reached. Consequently, a poor quality estimate of will be obtained.

Method II

The equations,

can be rearranged as

where is the matrix written as a vector. For 4 points,

which has the form , with a matrix. The solution is the (one dimensional) kernel of .

For n > 4 real point correspondences, is a matrix, and there will not be a solution to . In this case, a sensible procedure is to again minimise the residuals. It can be shown that the vector that minimises subject to , is the eigenvector with least eigenvector of . In the case of n = 4 the least eigenvalue is zero, and the eigenvector is the kernel of .

next up previous
Next: Summary of Plane Up: No Title Previous: Example 2: Synthetic

Bob Fisher
Wed Apr 16 00:58:54 BST 1997