in reply to OT:Math problem: Grids and conical sections.
No one seems to be using that the grid is, in a sense, nice - people seem to solve for generic 4 points, or wrongly assume that a square will project into a square again
The 4 points are in one plane, and on a cone - ie, they lie on a conic section. Since we only have a half of the full double cone, then a square(or even a rectangle) can only be inscribed in an ellipse, and not in a parabola or a half of a hyperbola.The only way to inscribe a rectangle in an ellipse is if their centres concide and the axes of that ellipse are parallel to the sides of the rectangle. But the cone vertex will always project on the major ellipse axis in the plane of the ellipse. Therefore, it would project somewhere on one of the 2 lines through the rectangle centre parallel to the sides - it can't go anywhere else.
Now follows a question - the height which we have measured at the 4 points - is it along the cone axis (ie, is the cone straight up), or is it in a random other direction? If the former, then the heights of 2 of the grid points will always be equal to the heights of the other 2. Once you find those matching pairs, then you know on which of the 2 lines to look for the cone vertex.I think if you sample 2 more points on that line, you should have 2 lines which intersect in the cone vertex.
If the height is in a random direction, then things become hairy.. let me work a bit more on that case.. update: the section is, indeed, conic, not conical...What a giant load of crap... For some reason I decided that the grid points themselves lie on the cone.. Whereas they clearly don't..
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