Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider an ellipsoid of semi-axes a, b, c (possibly prolate, b=c). I am interested in the "shadow" of this solid onto a distant plane, in a given direction d=(k,l,m) orthogonal to that plane. By shadow I mean the projected area onto the plane: each point on the surface of the ellipsoid is translated in the same direction until it intersects with a plane normal to it; the shadow is defined by the envelope of the intersection points.

First question: Is the projected curve an ellipse? (and what is its equation in terms of a,b,c and the direction vector d)?

Second question: What is the mean area of this shadow when averaged over all orientations of the ellipsoid (or, equivalently, the plane of projection)

I'm guessing this problem has been solved in the past; any references would be very welcome.



share|cite|improve this question
Have you seen this? – J. M. Dec 20 '10 at 13:17
See in particular pages 239-241 of that link. – Alex B. Dec 20 '10 at 14:37
Thanks, that's close. However I'm looking for an analytical solution. In fact, the statement that the projection is an ellipse is still not clear to me. – baptiste Dec 20 '10 at 16:24
So, if I follow you, it is obvious that the points of E that form the boundary of the projection are in a plane. But, why is this obvious? – to me it isn't. – baptiste Dec 20 '10 at 21:03
The points on $E$ that form the shadow border are characterized by the fact that there the gradient of the function $f(x,y,z):=x^2/a^2 + y^2/b^2+z^2/c^2-1$ is orthogonal to the direction ${\bf n}$ of the projection. Writing this out one finds that these points satisfy a linear equation, whence lie on a plane (through the origin). This plane intersects $E$ in an ellipse, and the latter projects to an ellipse. – Christian Blatter Dec 21 '10 at 9:14
up vote 4 down vote accepted

When a surface element $d\omega$ at a point $P$ in three-space is projected orthogonally into all possible directions then the average area of the image is $\rho\thinspace d\omega$ where $\rho$ is a universal constant. To determine the value of $\rho$ we argue as follows: The unit sphere $S^2$ has a surface area of $4\pi$, and the area of its projection is $2\pi$ in all directions (note that the "shadow" is doubly covered); whence also the average area of the projection is $2\pi$. So $\rho$ must be ${1\over2}$. What we have found out is valid for any surface element $d\omega$ of your ellipsoid $E$. It follows that the average area of the shadow of $E$ is equal to $\omega(E)/4$, where $\omega(E)$ is the total surface area of $E$ and we have taken into account that the shadow is covered twice. The value so obtained can be expressed as an elliptic integral in terms of $a$, $b$, $c$.

share|cite|improve this answer
Sounds good, thanks. A few questions though: - $\omega(E)$ would be the surface area of the ellipsoid? - why the factor of $1/4$ and not $1/2$? – baptiste Dec 20 '10 at 16:29
OK for the 1/4, however I fail to see why "This implies that any surface element ..." – baptiste Dec 20 '10 at 16:35
@ baptiste: It is obvious that any surface element $d\omega$ has an average projection size $\rho d\omega$ for some universal constant $\rho$. The first part of my argument serves to compute $\rho$ (=1/2) without actually doing the necessary integration. – Christian Blatter Dec 20 '10 at 20:32
OK – sorry for being dense – why is this obvious? ;) thanks again – baptiste Dec 20 '10 at 20:43
I think I understand most of the argument, except for the first part: how can we justify the first sentence (where $\rho$ is stated to be a universal constant)? – baptiste Dec 21 '10 at 10:33

I know this question has been answered long ago, but I wanted to provide you with a decent paper (albeit astrophysical in nature), since it does deal with the projections of tri-axial ellipsoids (with arbitrary semi-axes, and orientation) onto a plane. Binggeli et al. 1980.

In summary: The answer to your first question is yes. Quantities constant on ellipsoids (e.g. - isodensities (same mass) or isophotes (same light) ) project to ellipses.

The perceived axis ratio (minor to major) of your new ellipse will be the following:

$$Q = \sqrt{\frac{j+l - \sqrt{(j-l)^{2}+4k^{2}}}{j+l+\sqrt{(j-l)^{2}+4k^{2}}}} $$

where j,k, and l are geometric functions dependent upon the instrinsic axis ratios, and orientation angles with respect to the line of sight of your ellipsoid. $$j = q^{2}\sin^{2}\theta+p^{2}\sin^{2}\phi\cos^{2}\theta+\cos^{2}\phi\cos^{2}\theta $$ $$l = p^{2}\cos^{2}\phi+\sin^{2}\phi $$ $$k=(1-p^{2})\sin\phi\cos\phi\cos\theta $$

Lots of transcription here - I'd definitely double-check this in the paper. p is the intermediate-to-major axis ratio, q is the minor-to-major axis ratio, and theta and phi define the observer's orientation relative to the intrinsic coordinate system of the ellipsoid (meaning the coordinate system is aligned along the semi-axes).

A few other notes to avoid confusion with the notation: 1) Prolate spheroids mean p=q, and 2) Oblate spheroids mean p=1, with q less than p in value.

Sorry that this is a little more physics-y than math-y. I understand this is a mathematics stack exchange site. I hope this helps!

share|cite|improve this answer
nice reference, thanks! – baptiste Aug 24 '13 at 16:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.