Triangulation solving 3D coordinate-direct linear conversion (Linear TRANSFORMATION-DLT) algorithm

In a long vision of 3D reconstruction, when finding multiple frames of matching points and corresponding camera gestures, we need to find the matching points x and X ' and corresponding p and P ' I

They get x = PX, x ' = P ' x.

We get the two equations above, how to solve the equations.

Here we go back to the 2D projection transformation to say, for the 2D plane on the two corresponding point <x, X ', we can find a single-should matrix H to establish the relationship between the X ' =

Hx. H is a 3x3 matrix, so x and X ' are 2D coordinates in homogeneous coordinates, their order of magnitude is different, then X ' = Hx is not necessarily true, but satisfies

Cross product relationship x ' x Hx = 0. Obviously, X and X '

In the same direction, the cross product is 0). At this point, set

So, x ' x Hx = 0 satisfies

Because, so we can write the form in AX = 0

If we use XI ' multiplied by the first line, Yi ' multiplies the second line and adds it, it gets the third row, so only the first two lines are linearly independent. So the final A is the 2x9 matrix:

The third dimension W of the homogeneous coordinate system can be set to 1. Then a set of matching points can get two equations.

There are three types of solutions for a:

1. Give 4 match points to get a 8X9 formula, and rank (A) = 8. So the null-space of a is a solution.

2. When there are more than 4 points, we need to solve the super-definite solution. The solution we want is the singular vector corresponding to the smallest singular value of a.

3. Since the solution of H depends on a coefficient scale, we can constrain h by changing the coefficients. If we make h3,3 1, then we get

For n points, our solution to a 2nx9 matrix A, h is the singular vector corresponding to the smallest singular value.

So back to our triangulation problem, we found that two formulas x ' = PX and X ' = Hx are very similar. So x x (PX) = 0, we can get ax=0, where A is

The solution is the same as above.

Since I am also a beginner, I am interested in solving AX = 0. It says that the solution of the least squares of X is why the singular vectors corresponding to the svd least singular values.

For AX = 0 We are interested in the non-0 solution of X. If X is the solution then KX is also the solution. In order to make the solution unique. We join the constraint.

To a do SVD decomposition,. Our goal is to minimize, that is. Since both U and V are unit vectors,

And, we make, then, we finally optimize. In this case, D is a diagonal matrix with singular values stored diagonally and descending from top to bottom. So this solution is y = (0,0,..., 1) '; so x = Vy. Which is the most vector of the singular vector matrix.