
Re: 3D geometry problem
Posted:
Apr 2, 2012 10:29 AM


> 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 xy 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

