Two View Triangulation of 3D Skew Lines


Assume that we have parametric lines and , the challenge is to find the points and of closest approach. First, we must test the assumption that our lines are skew, meaining they are not parallel and do not intersect. To frame this, we take the forms of and and simplify them by thinking of them as parametic vectors where and , represent the camera position vectors and and represent the vector was previously calculated from the subtraction of match coordinates with the camera vector. We make the simple equations:

To make sure that the lines are not parallel, which is unlikely, we must verify that their cross product is not zero. if then we have a degenerate case with infinitely many solutions. As long as we know this is not the case we can proceed. We know that the cross product of the two vectors is perpendicular to the lines and . We know that the plane , formed by the translation of along , contains . We also know that the point is perpendicular to the vector . Thus, the intersection of with is also the point, , that is nearest to , given by the equation:

This also holds for the second line , the point , and vector . with the equation:

Now, given to points that represent the closest points of approach, we simply find the midpoint :











This is copied from a section of my thesis. If you found this useful to your research please consider using the following bibtex:

@mastersthesis{CalebAdamsMSThesis,
  author={Caleb Ashmore Adams},
  title={High Performance Computation with Small Satellites and Small Satellite Swarms for 3D Reconstruction},
  school={The University of Georgia},
  url={http://piepieninja.github.io/research-papers/thesis.pdf},
  year=2020,
  month=may
}