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Re: 3D geometry problem
Posted:
Apr 2, 2012 10:29 AM
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> Hello Robert,
>
> By observer I mean the virtual camera that renders a
> 3D scene. It can view the plan from any angle but at
> any time I know the plane equation, the camera point
> of view vector (or its parametric equation) thus I
> know the point B and the angle beta.
>
> What I need to do is given any point A on the plane,
> to find D with the following characteristics:
> - D must be on the camera pov line
> - the angle alpha must be at a specific value (which
> never changes)
>
> So mathematically I think I need to express D(x,y,z)
> coordinates in terms of A,B,alpha or beta.
>
> I also need C but I can find it from D and the line
> coordinate. And the inverse is also possible so if
> it's easier a solution expressing C in term of
> A,B,alpha or beta is also valid.
>
Unless there is a variable you haven't explained, it's not possible in general. I see four equations but only three variables.
I'm assuming the plane is the x-y plane. D is (x,y,z). Your observer's eye is at a known point (v1, v2, v3). B is (b1,b2,0). A is (a1,a2,0). From the tangent of the angles alpha and beta we immediately get two equations relating x, y, and z. (v1,v2,v3) lies on the vector from B to D, so we get two more equations. That's four equations in three variables, no good. There are seven parameters, a_i, b_i, v_i. Since these are polynomials, one could compute the resultant of the four equations, which is a relation that must be satisfied among the parameters to make it possible.
Robert H. Lewis
Fordham University
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