Timeline for Solving 3D equation system (inverse-projecting a triangle)
Current License: CC BY-SA 3.0
7 events
when toggle format | what | by | license | comment | |
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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
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Oct 17, 2012 at 11:45 | history | asked | AndresN | CC BY-SA 3.0 |