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Lecture Planar Scenes

Reading Ma 2004:Ch 5

Last weeks exercise

Not designed as an exercise to check that you have learnt what I know.
- Rather, it is designed as an experiment, as I would use it to test my own understanding.
It sets up a closed loop.
- the final result can be checked against the original data
It demonstrates the eight-point algorithm, but it also demonstrates image capture (projection).
- but this requires that you take the time to comprehend each step …

If you can complete and comprehend all the steps, you have understood the core of 3D reconstruction …

however, there is more

A plane \(P\) is described by an equation \[N^T{X}=d\]
- where \(N=(n_1,n_2,n_3)\) is a vector orthogonal on \(P\)
Consider object points \(X_1,X_2,\ldots,X_n\in P\).
- they all satisfy \(N^TX_i = d\)
- or \(\frac1dN^TX=1\) (1)
Extra constraint compared to the case for the eight-point algorithm
Consider the transformation between camera frames \[X'=RX+T\]
inserting from (1), we have \[X'=RX+T\frac1dN^TX=(R+T\frac1dN^T)X=HX\]
- where \(H=R+T\frac1dN^T\)
- \(H\) depends on \((R,T)\) as well as \((N,d)\).
Consider the image points \(x'=X'/\lambda'\) and \(x=X/\lambda\).
- we get \(x'\sim Hx\) (planar) homography
multiplying both sides by \(\hat x'\), we get
Planar epipolar constraint \[\hat x'Hx=0\]
because \(x'\sim Hx\), for any \(u\in\mathbb{R}^3\), \(u\times x'=\hat ux'\bot Hx\)
hence \(x'^T\hat u Hx=0\) for all \(u\in\mathbb{R}^3\)
thus the epipolar constraint is under-defined
it follows that the eight-point algorithm cannot work

An image point \(x\) corresponding to \(p\in P\) uniquely determines \(x'\sim Hx\)
- if \(p\not\in P\), \(x'\) only ends up on the epipolar line