Handling CSDN self https://blog.csdn.net/u013213111/article/details/101233824
1 DoF of Homography
Reference:
Nanjing Higher geometry class 19 Normal 25 '
Tracing the source, or go higher geometry / analytic geometry of it.
For a two-dimensional projective homography 3 * 3 matrix
\ [H = \ left [\ begin {matrix} h_ {11} & h_ {12} & h_ {13} \\ h_ {21} & h_ {22} & h_ { 23} \\ h_ {31} &
h_ {32} & h_ {33} \ end {matrix} \ right] \] Although there are nine elements, but does not require a specific value of each element, you just need to proportional relationship to these elements, to get proportional relationship to determine a homography, therefore minus one degree of freedom.
So if \ (33 is H_ {} \ NEQ 0 \) , A \ (H \) for method parameter is normalized by dividing each element may be \ (} 33 is H_ {\) , after such treatment , the original \ (h_ {33} \) becomes 1, and the remaining eight unknown ratio, as long as the ratio solved to eight, i.e., 8 degrees of freedom. In fact, in theory, you do not have to use \ (H_ {33} \) , although this is generally used, after all, quite possible for other elements of 0.
BTW, What IF \ (= 0} 33 is H_ {\) ? If the listed equation, it can be found to be a \ ((0,0, w) \ ) point (i.e. the origin) is mapped to \ ((x, y, 0 ) \) point (i.e. the point at infinity) projective. However, I am afraid I stupid, and can not imagine it would be a kind of projection.
In addition to the \ (h_ {33} \) as a "reference value" for normalizing, there is a way \ (\ sum h_ {ij} ^ 2 = 1 \) as a constraint, or is \ ((\ sum h_ {ij} ^ 2) ^ {1/2} = 1 \) as a constraint, the matrix is also called L2 norm or Euclidean norm oR Frobenius norm. This is the "length" of the matrix is normalized as a reference value.
Since the DoF Homography 8, so that four pairs of points can be solved with a Homography.
Back to higher geometry, there is such a theorem:
Corresponding to a two-dimensional projection may be any of the known four pairs uniquely determined corresponding points (three points for each side four collinear points no).
Therefore, the use of four pairs of points in computer vision to determine a homography in mathematics is based on!
As a perfectionist, obsessive-compulsive disorder patients expressed very satisfied with this result.
2 parallel lines and the point at infinity
Reference:
Projective Transformations
Suppose \ (w = 1 \) is a projective plane, \ (AX + by + c_1 and = 0 \) and \ (ax + by + c_2 = 0 \) two parallel lines on the plane, how understand the intersection of these two parallel lines are infinite point of it?
Noted that each point in a straight line through both the center of projection of the 3D space corresponding to a projective plane, the two parallel lines which will correspond to the two planes in 3D space, while the line of intersection of two planes , i.e. \ (w = 0 \) a straight line on this plane \ (AX + by = 0 \) , according to the definition of projective geometry, and this line will correspond to the point at infinity.