A web page on "Projective Bases for the Projective Plane" attributed to Bill Triggs says, "Any four distinct coplanar points (no three of which are colinear) form a projective basis for the plane they span. Given four such points A,B,C,D, a fifth point M in the plane can be characterized as the intersection of one line of the pencil through D with one line of the pencil through B. ... Given four known points on a 3D plane and their perspective images, a fifth unknown point on the plane can be reconstructed from its image by [expressing the image in terms of the four known image points, and then using the same homogeneous planar coordinates (lambda, u, v) to recreate the corresponding 3D point.]
I don't understand how to actually do the part that I placed in square brackets. Could someone point out a more step-by-step example of what that means?
My understanding is that this is an alternative to first computing the homography and then applying it to map the fifth point, but I may be completely wrong.
I would solve the problem as follows. Let X[1], ... X[5] be the coplanar scene points and x'[1], ..., x'[5] be their images under projection. Let x[i] = [I 0] X[i], where I is identical 3x3 matrix. Compute an homography H for the 4 image correspondences x[i]x'[i] and find x[5] = H x'[5]. Then X[5] = [x[5] t]. The 3d plane pi (which is a 4-vector) is found from 4 linear equations pi^T X[i] = 0. Find t from pi^T X[5] = 0. I believe the approach with cross-ratios is just another way to get the solution as you supposed.
Thanks, I may have to do it that way. A couple papers on computer vision suggest that working directly with the cross ratios to map additional points and establish correspondences is faster than first calculating a putative homography. [I failed to get the question on the Computer Vision topic].
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