Correct projection of error ellipsoid onto horizontal plane.

Question

While solving some problem, I obtained the error ellipsoid as an uncertainty estimate of point location in 3-D space. In fact, error ellipsoid is given by standard error (SERR), azimuth, and dip of three orthogonal principal vectors that fully describes an ellipsoid. For example, we can think of a case where 3 vectors are $(SERR, AZ, DIP)=(2.04,\,173,\,17),\,(1.45,\,323,\,69),\,(0.71,\,80,\,9)$. Here, the unit of $SERR$ is in km and unit of $AZ$ and $DIP$ are both in degrees.

Actually, what I want to do is not plotting error ellipsoid in 3-D space but plotting projection of ellipsoid (i.e. error ellipse) onto horizontal (x-y) plane. To do this, I thought of the method of finding major axis in horizontal plane by calculating horizontal projection of each principal axes by multiplying $\cos(DIP)$ to each $SERR$ and choosing the one with maximum projection length as major axis and taking corresponding azimuth for projection. However, I also should determine the minor axis in horizontal plane but actually azimuths of principal axes are not exactly orthogonal in 3-D so I got confused about way to get correct minor axis in this way. Is there more appropriate way of projecting ellipsoid onto horizontal plane?

Answer 1

If $v_1$, $v_2$ and $v_3$ are the orthogonal principal vectors of the ellipsoid, let $A$ be the matrix having vectors $$ \pmatrix{\displaystyle{v_1\over|v_1|^2} &\displaystyle {v_2\over|v_2|^2} &\displaystyle{v_3\over|v_3|^2}} $$ as columns, and define $M=AA^T$. Then the cartesian equation of the ellipsoid can be written as $$ \tag1 \pmatrix{x & y & z}M\pmatrix{x\\ y\\ z}=1. $$ The gradient $n=M\pmatrix{x\\ y\\ z}$ of the left hand side gives the direction of the normal: the points corresponding to the border of the projection on plane $xy$ (which is an ellipse) are those whose gradient has vanishing $z$ component, i.e. those on the plane described by $n_z=0$.

Rewrite the equation of that plane, expressing $z$ as a function of $x$, $y$ and plug the result into $(1)$: the resulting equation is that of the desired projection.

Example.

Suppose $v_1=(1,1,2)$, $v_2=(2,0,1)$, $v_3=(-1,5,-2)$. Then we get: $$ m=\left( \begin{array}{ccc} \frac{17}{90} & \frac{1}{45} & -\frac{1}{45} \\ \frac{1}{45} & \frac{1}{18} & \frac{2}{45} \\ -\frac{1}{45} & \frac{2}{45} & \frac{7}{45} \\ \end{array} \right) $$ and the equation of the ellipsoid is $$ \tag2 \frac{1}{90} \left(17 x^2+4 x (y-z)+5 y^2+8 y z+14 z^2\right)=1. $$ We also get $$ n=\pmatrix{\frac{17 x}{90}+\frac{y}{45}-\frac{z}{45}\\ \frac{x}{45}+\frac{y}{18}+\frac{2z}{45}\\ -\frac{x}{45}+\frac{2 y}{45}+\frac{7 z}{45}} $$ and from $\displaystyle n_z=-\frac{x}{45}+\frac{2 y}{45}+\frac{7 z}{45}=0$ we obtain $$ z={1\over7}(x-2y). $$ Plugging this into $(2)$ we finally obtain the equation of the projection: $$ 13x^2+4xy+3y^2=70. $$