next up previous contents
Next: Conclusion Up: Multi view geometry Previous: Three view geometry   Contents


Multi view geometry

Many people have been studying multi view relationships [59,156,34]. Without going into detail we would like to give some intuitive insights to the reader. For a more in depth discussion the reader is referred to [88].

An image point has 2 degrees of freedom. But $n$ images of a 3D point do not have $2n$ degrees of freedom, but only 3. So, there must be $2n-3$ independent constraints between them. For lines, which also have 2 degrees of freedom in the image, but 4 in 3D space, $n$ images of a line must satisfy $2n-4$ constraints.

Some more properties of these constraints are explained here. A line can be back-projected into space linearly (3.9). A point can be seen as the intersection of two lines. To correspond to a real point or line the planes resulting from the backprojection must all intersect in a single point or line. This is easily expressed in terms of determinants, i.e. $\vert {\tt\Pi}_1 {\tt\Pi}_2 {\tt\Pi}_3 {\tt\Pi}_4 \vert=0$ for points and that all the $3 \times 3$ subdeterminants of $[{\tt\Pi}_1 {\tt\Pi}_2 {\tt\Pi}_3]$ should be zero for lines. This explains why the constraints are multi linear, since this is a property of columns of a determinant. In addition no constraints combining more than 4 images exist, since with 4-vectors (i.e. the representation of the planes) maximum $4 \times 4$ determinants can be obtained. The twofocal (i.e. the fundamental matrix) and the trifocal tensors have been discussed in the previous paragraphs, recently Hartley [54] proposed an algorithm for the practical computation of the quadrifocal tensor.


next up previous contents
Next: Conclusion Up: Multi view geometry Previous: Three view geometry   Contents
Marc Pollefeys 2002-11-22