Shadow of an ellipsoid - Uniform Direction Light

By checking $\vec{u_i}\cdot\vec{u_j}=\delta_{ij}$ , it is easily verified that $(\vec{u_1},\vec{u_2},\vec{u_3})$ form an orthonormal basis. It will be convenient to work in this basis, because it is aligned with the ellipsoid's axes. Let's put our origin at the centre of the ellipsoid. If the light rays have direction $\vec{r}$ , this vector can be expressed in the u-basis as $\vec{r}\cdot \vec{u_1}, \vec{r} \cdot \vec{u_2}, \vec{r} \cdot \vec{u_3} = \frac{1}{9}(8, 13, -35)$ , or, because the length does not matter, just $\vec{r}=(8,13,-35)$

Consider the function $f(\vec{u})= 4u_1^2 + u_2^2+ 9u_3^2$ then our ellipsoid is the surface where $f(\vec{u})=900$ . The gradient $\nabla f$ is orthogonal to the surface, so $\nabla f =(\frac{\partial f}{\partial u_1}, \frac{\partial f}{\partial u_2}, \frac{\partial f}{\partial u_3}) = (8u_1, 2u_2, 18u_3)$ , so that a normalvector to the surface at a point $\vec{u}$ on the surface is given by $\vec{n}=( 4u_1, u_2, 9u_3)$

We are interested in the light rays that are tangent to the ellipsoid. These rays have to be normal to $\vec{ n}$ , so we demand $\vec{r} \cdot \vec{n}=0$ or, expressing the dot product in coordinates: $32u_1 + 13u_2 - 315u_3=0$

This linear equation defines a plane, that I will call P, and allows us to express $u_3^2 = \frac{(1024u_1^2+832u_1u_2y +169u_2^2)}{315^2}$ so that the quadratic equation $4u_1^2 + u_2^2 + 9u_3^2 = 900$ can be written as $4u_1^2 + u_2^2 + \frac{9}{315^2}(1024u_1^2+832u_1u_2 +169u_2^2)  = 900$ or $45124u_1^2 + 832u_1u_2 + 11194u_2^2 = 9922500$

This ellipse can be rotated in the $u_1$ - $u_2$ -plane to get it into the standard form, with half axes $14.82...$ and $29.77...$ and has area $1387.231501...$ .

Note that this ellipse is just the projection of the actual ellipse, that lives in plane P, onto the $u_1$ - $u_2$ -plane. The actual ellipse is bigger because of the inclination of the plane P. Vectors in the plane P are $(315,0,32)$ and $(0 315, 13)$ so a normal vector to the plane is $\vec{v}=(32,13,-315)$ Hence the true area of the ellipse in plane P itself is bigger by a factor $\frac{|v|}{|v_3|}=\frac{\sqrt{100418}}{315}=1.00599...$ . Its area is $1395.546...$ (actually quite close to the maximum cross section of $450π=1413.7...$ ).

Further note that the light rays are not perpendicular to plane P. The cross section of the shadow (perpendicular to the light bundle) is smaller by a factor $\frac{\vec{r}\cdot\vec{v}}{|r||v|} = \frac{11450}{\sqrt{1458\cdot100418}} = 0.9462828...$

Finally, the lightbundle makes an angle with the z-axis, which makes the shadow on the x-y plane bigger by a factor $\frac{\sqrt{1^2+1^2+4^2}}{4}=1.06066...$

The combined effect of these three projections is a factor $\frac{1145}{1134}=1.009700176...$ so that the shadow on the xy-plane has $\text{Area}= 1387.231501..×1.009700176..=\boxed{1400.68789}$

If $(x', y', z')$ are the coordinates of a point, on the ellipsoid, with respect to $\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}$ , then we know that,

$\dfrac{ {x'}^2 }{15^2} + \dfrac{ {y'}^2 }{30^2} +\dfrac{ {z'}^2 }{10^2} = 1$

Thus if we define $\mathbf{r'} = (x', y', z')$ , and $\mathbf{r} = (x, y, z )$ ,then

${\mathbf{r'}}^T D_e \mathbf{r'} = 1$ ,

where $D_e$ is a diagonal matrix whose diagonal entries are, in order, $\dfrac{1}{15^2}$ , $\dfrac{1}{30^2}$ and $\dfrac{1}{10^2}$ . The relationship between $\mathbf{r'}$ and $\mathbf{r}$ is simply,

$\mathbf{r} = \mathbf{r_0} + R_e \mathbf{r'}$ where $R_e = [ \mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3} ]$ , and $\mathbf{r_0}$ is the position vector of the center of the ellipsoid.

The three columns of $R_e$ are orthonormal, hence ${R_e}^{-1} = {R_e}^T$ , and thus, the equation of the ellipsoid is,

$({\mathbf{r}-\mathbf{r_0}})^T Q_e (\mathbf{r}-\mathbf{r_0}) = 1$ , where $Q_e = R_e D_e {R_e}^T$

First thing we need to do is find the coordinates of the center of the ellipsoid, which is the vector $\mathbf{r_0}$ . To that end, we know that at the tangency point (which is given by $\mathbf{r} = (0,0,0)$ ), the normal vector to the surface of the ellipsoid is pointing in the negative $\mathbf{k}$ direction, where $\mathbf{k} = (0, 0, 1)$ . Now, the normal vector is parallel to $Q_e (\mathbf{r}-\mathbf{r_0} ) = - Q_e \mathbf{r_0}$ , therefore, we have, $Q_e \mathbf{r_0} = \alpha \mathbf{k}$ for some positive constant $\alpha$ . From which, $\mathbf{r_0} = \alpha {Q_e}^{-1} \mathbf{k}$ . Plugging this, and $\mathbf{r} = \mathbf{0}$ , into the ellipsoid equation, we obtain,

$\alpha = \dfrac{1}{\sqrt{ \mathbf{k}^T {Q_e}^{-1} \mathbf{k} }}$

Hence,

$\mathbf{r_0} = \dfrac{1}{\sqrt{ \mathbf{k}^T {Q_e}^{-1} \mathbf{k} }} {Q_e}^{-1} \mathbf{k}$

So, now the ellipsoid is completely specified.

The shadow of the ellipsoid on the $xy$ plane is generated by the intersection of the $xy$ plane and the non-uniformly scaled cylinder that is tangent to the ellipsoid, and which has its axis parallel to the fixed direction of light rays $\mathbf{d_0} = (1, -1, -4)$ .

Tangency points on the ellipsoid are characterized by the vector equation,

$\mathbf{d_0}^T Q_e (\mathbf{r} - \mathbf{r_0} ) = 0$

This is an equation of a plane that passes through the center of the ellipsoid. Intersecting a plane with an ellipsoid was addressed here . The resulting ellipse can be parameterized as,

$\mathbf{r_e}(t) = \mathbf{r_0} + \mathbf{v_1} \cos t + \mathbf{v_2} \sin t$

where $t \in \mathbb{R}$ . Points on the surface of the tangent cylinder to the ellipsoid can be expressed as,

$\mathbf{r_c}(t) = \mathbf{r_e}(t) + s \mathbf{d_0}$

where $t \in \mathbb{R}$ .

The above is a vector equation of the lateral surface of the cylinder. To obtain an algebraic equation, we note that,

$\mathbf{r_c} - \mathbf{r_0} = A \mathbf{w}$

where,

$A = [ \mathbf{v_1} , \mathbf{v_2}, \mathbf{d_0} ]$ , and, $\mathbf{w} = \begin{bmatrix} \cos t \\ \sin t \\ s \end{bmatrix}$

Now, let

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

Then, $\mathbf{w}^T Q_0 \mathbf{w} = 1$

But $\mathbf{w} = A^{-1} (\mathbf{r_c} - \mathbf{r_0} )$

Hence, the algebraic equation of the shadow cylinder is

$(\mathbf{r_c} - \mathbf{r_0} )^T Q_c (\mathbf{r_c} - \mathbf{r_0} ) = 1 \hspace{30pt} (1)$

where $Q_c = A^{-T} Q_0 A^{-1}$

The final step is to intersect this cylinder with the $xy$ plane. This is discussed below.

A point $\mathbf{r}$ on a plane is given by,

$\mathbf{r} = \mathbf{r_1} + W \mathbf{u} \hspace{30pt} (2)$

where $\mathbf{r_1}$ is any arbitrary point on the plane, and $W = [ \mathbf{w_1} , \mathbf{w_2} ]$ is a $3 \times 2$ matrix whose columns are arbitrary unit vectors that together with the normal to the plane form an orthonormal basis for $\mathbb{R}^3$ .

In the case of the $xy$ -plane, it is natural to choose $\mathbf{r_1} = (0, 0, 0)$ , $\mathbf{w_1} = (1, 0, 0)$ and $\mathbf{w_2} = (0, 1, 0)$ .

Substituting this vector expression (2) into the equation of the cylinder (1), one obtains,

$( \mathbf{r_1} - \mathbf{r_0} + W \mathbf{u} )^T Q_c (\mathbf{r_1} - \mathbf{r_0} + W \mathbf{u}) = 1$

Expanding this quadratic form gives,

$\mathbf{u}^T (W^T Q_c W) \mathbf{u} + 2 \mathbf{u}^T W^T Q_c (\mathbf{r_1} - \mathbf{r_0}) + (\mathbf{r_1} - \mathbf{r_0})^T Q_c ( \mathbf{r_1} - \mathbf{r_0}) = 1$

Take $\mathbf{u_0} = - (W^T Q_c W)^{-1} W^T Q_c (\mathbf{r_1} - \mathbf{r_0} )$ , then the above equation becomes,

$(\mathbf{u} - \mathbf{u_0})^T (W^T Q_c W) (\mathbf{u} - \mathbf{u_0}) = C$

where the constant $C$ is given by $C = 1 - (\mathbf{r_1} - \mathbf{r_0})^T Q_c (\mathbf{r_1} - \mathbf{r_0}) + {\mathbf{u_0}}^T (W^T Q_c W) \mathbf{u_0}$

Dividing by $C$ , we obtain,

$(\mathbf{u} - \mathbf{u_0})^T Q_1 (\mathbf{u} - \mathbf{u_0}) = 1$

where $Q_1 = \dfrac{1}{C} (W^T Q_c W)$ .

The final step is to diagonalize $Q_1$ into $Q_1 = R D R^T$ . The diagonal elements of the diagonal matrix $D$ are the eigenvalues of $Q_1$ which are the reciprocal of the square of the semi-major and semi-minor axes lengths $a$ and $b$ of the resulting ellipse of intersection.

Hence, at last the shadow area is given by $\pi a b$ .