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Oct 22, 2012 at 9:40 comment added AndresN this is perhaps the only realistic way indeed, thanks!
Oct 18, 2012 at 15:10 comment added J.C. Ottem AndresN, did you try Newton's method? Since the equations are quadrics, this should be pretty straightforward to implement and the iteration should converge pretty fast.
Oct 18, 2012 at 14:12 comment added AndresN Thanks Daniel and Mr Ottem! I can see the elimination works (tried with Maple). Solving it takes huge amount of steps, so I can't use it in practice most likely (will lose the precision long before done), but this is my intro to quadrics I guess. PS I can assume (x, y, z) are all positive - they are z-distances from camera to 3 points
Oct 17, 2012 at 21:52 comment added J.C. Ottem The system does have solution in terms of radicals. Note that if $(x,y,z)$ is a solution, then $(-x,-y,-z)$ is also a solution. This means that if you look at the equations $(Ax-By)^{2}+(Cx-Dy)^{2}+(x-y)^{2}-G=0$ $(Ax-Ez)^{2}+(Cx-Fz)^{2}+(x-z)^{2}-H=0$ $(Ez-By)^{2}+(Fz-Dy)^{2}+(z-y)^{2}-I=0$ and elimiate the variables $y$ and $z$, this is going to give you a polynomial of degree 4 in $x^2$. Likewise, eliminating $x$ and $z$ (resp. $x$ and $y$) will give you a polynomial of degree 4 in $y^2$ (resp. degree 2 in $z^2$). Maple and Mathematica or any CAS with Grobner basis can do this easily.
Oct 17, 2012 at 13:24 comment added Daniel Loughran I'm not sure such equations have a name, other than "intersections of three quadrics". For generic $A,B,\ldots,I$, this will have only finitely many solutions. It fact it will have at most $8$ solutions (there might actually be less due to multiple solutions or solutions at infinity). Have you tried calculating a Gröbner basis? It might do the trick.
Oct 17, 2012 at 13:05 history edited AndresN CC BY-SA 3.0
corrected the number expected of answers from 1 to 2
Oct 17, 2012 at 11:45 history asked AndresN CC BY-SA 3.0