Find the projected ellipse on a plane from a given $3D$ ellipse

44 Views Asked by At

Given the ellipse

$ E(t) = C + V_1 \cos t + V_2 \sin t $

I want to generate rays from a point $P$ lying outside the plane of $E(t)$, then intersect those rays with the projection plane $n^T (r - r_0) = 0 $, thus generating a projected ellipse. What I want is to find the direction and length of the semi-axes of the projected ellipse.

My approach:

The equation of the lines connecting $P$ to $E(t)$ is

$ r(t) = P + \lambda ( E(t) - P ) , \lambda \in \mathbb{R}$

So,

$ r - P = \lambda ( C - P + V_1 \cos t + V_2 \sin t ) $

Define the matrix

$ G = [V_1 , V_2, C - P ] $

then

$ r - P = \lambda G u $

where $ u = [\cos t , \sin t , 1 ]^T $

This vector satisfies

$ u^T Q_0 u = 0 $

where

$ Q_0 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && - 1 \end{bmatrix} $

Hence, we now have the equation of all the rays as the cone

$(r - P) G^{-T} Q_0 G^{-1} (r - P) = 0 $

Intersecting this cone with the plane $n^T (r - r_0) = 0 $ gives the desired projection ellipse (it could also a parabola or a hyperbola).

Your comments, alternative solutions, are highly appreciated.