Computing 3D-homography with 5 3D-points

Question

I've got a set of 3D-points in a projective space and I want to transform them into a metric 3D space so that I could measure distances in meters.

In order to do so, I need a 3D to 3D homography, which is a 4x4 matrix with 15 degrees of freedom (so I need 5 3D-points to get 15 equations). I have a set of these 5 3D-points from the projective space and their corresponding 5 3D-points aligned in the metric space (which I expect the 5 projective points to be transformed to).

I can't figure out how to estimate the homography matrix. At first I tried:

A=np.vstack([p1101.T, p1111.T, p0101.T, p0001.T, p0011.T])
b=np.array([[1,1,0,1], [1,1,1,1], [0,1,0,1], [0,0,0,1], [0,0,1,1]])
x, _, _, _ = np.linalg.lstsq(A,b)
H = x.T

where p1101 is a [X,Y,Z,1] point which corresponds to [1,1,0,1] in the 3D metric space, etc.. However, this is not correct since I'm in projective space, so I need to create somehow an equation set where I divide the rows of H with its last or something like that.

I thought maybe there is an implemented method that will do it for me, for example in opencv, but didn't find. Any help would be appreciated.

user_6183071 · Answer

I finally solved this question with a friend, and would like to share the solution.

Since in projective space, one needs to solve an equation set where the homogene coordinate of the outcome is the denominator of each other coordinate. i.e, if you want to find a 4x4 homography matrix H, and you have matching 3D points x and b (b is in the meteric space), you'll need to optimize the search of H parameters such that H applied on x will give a vector v with 4 coordinates, such that all the first three coordinates of v divided by the last coordinate are b. written in numpy:

v = H.dot(x)
v = v[:3]/v[3]
v == b  # True

mathematically, the optimization is based on this (this is focused on the first coordinate only, for simplicity, but other coordinates are done the same way):

so in python one needs to arrange the equations for the solver in the explained manner, with 5 matching points. The way that was purposed in the question is good (just didn't solve the right problem), and in these terms it will make Ax=b least squares optimization such that A is 15x15 matrix, and b is a 15 dimensional vector. Each matching point generates 3 equations, then 5 matching points will generate 15 equations built into the matrix A, thus solving the 15 DOF of the 3D homography H.

Computing 3D-homography with 5 3D-points

1 Answers