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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.

Thanks,

baptiste

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    Have you seen [this](http://books.google.com/books?id=FretKHY7xSMC&pg=PA233)?2010-12-20
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    See in particular pages 239-241 of that link.2010-12-20
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    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.2010-12-20
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    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.2010-12-20
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    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.2010-12-21

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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 $a$, $b$, $c$.

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    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 $1/2$?2010-12-20
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    OK for the 1/4, however I fail to see why "This implies that any surface element ..."2010-12-20
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    @ 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.2010-12-20
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    OK – sorry for being dense – why is this obvious? ;) thanks again2010-12-20
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    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)?2010-12-21
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    Assume $d\omega$ orthogonal to the $z$-axis. Then the area of its projection along ${\bf n}$ is given by $|\cos\theta|\thinspace d\omega$ where $\theta$ is the angle between ${\bf n}$ and the $z$-axis. Now you have to average this when ${\bf n}$ runs over the 2-sphere.2010-12-21
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    ... which indeed gives 1/2. Thanks, I see now.2010-12-21
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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!

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    nice reference, thanks!2013-08-24