Suppose you have an ellipsoid with semi-axes lengths $a,b,c$, and you a rectangular box (i.e. cuboid) of dimensions $L, W, H$, with its length $L$ along the $x$ axis, its width $W$ along the $y$ axis, and its height $H$ along the $z$ axis. It can be shown that if the box fits the ellipsoid exactly, i.e. the ellipsoid is tangent to all $6$ sides of the box, then we must have

$$ \dfrac{1}{4} (L^2 + W^2 + H^2) = a^2 + b^2 + c^2 \hspace{20pt}(*) $$

It can also be shown that

$$ e_1^T R D R^T e_1 = \left( \dfrac{L}{2} \right)^2 \hspace{20pt}(1)$$

$$ e_2^T R D R^T e_2 = \left( \dfrac{W}{2} \right)^2 \hspace{20pt} (2)$$

$$ e_3^T R D R^T e_3 = \left( \dfrac{H}{2} \right)^2 \hspace{20pt} (3)$$

where $e_1, e_2, e_3$ are the coordinate unit vectors, and $R$ is the (rotation) matrix that describes the orientation of the semi-axes of the ellipsoid with respect to the world frame, where it is assumed that $L$ is along the $x$ axis, $W$ is along the $y$ axis, and $H$ is along the $z$ axis. Matrix $D$ is a diagonal matrix with the diagonal entries being $a^2, b^2, c^2 $.

Because of the relation $(*)$, only TWO equation of the set $(1), (2), (3)$ are independent, while the third follows from $(*)$.

Since $R$ can be parameterized using $3$ angles, and now we have two constraints on it, then it follows that there is an infinite number of rotation matrices $R$ that satisfy equations $(1), (2), (3) $. And it also follows that $R$ can be parameterized by a single parameter, i.e. we have a single degree of freedom for describing $R$.

My question, is how to find such parameterization of $R$ ?

  • $\begingroup$ Don't you think that $e_1,e_2,e_3$ should be chosen as unit vectors along the direction of the edges of the box ? $\endgroup$
    – Jean Marie
    Commented Jul 22, 2023 at 13:40
  • $\begingroup$ Yes. It is assumed that box is standing on the $xy$ plane with its height along the $z$ axis. But I will make this clear in my question. $\endgroup$ Commented Jul 22, 2023 at 14:04

1 Answer 1


I've found a parameterization using the tangency points, I hope it may be of help.

Let $P$, $Q$, $R$ be the tangency points between the ellipsoid and three concurrent faces of the cuboid (see figure below). It follows from the properties of the ellipse that the line joining the centre $O$ of the cuboid with the midpoint of $PQ$ must intersect the line of the common edge between the two faces where $P$ and $Q$ lie. If $M$ and $U$ are the projections of $P$ and $Q$ on that edge, then it is not difficult to prove that the above property entails $$ PM : H = QU : L $$ and of course two analogous proportions arise from the other pairs of tangency points.

enter image description here

Set up a coordinate system such that the cuboid is described by $$ -{L\over2}\le x\le{L\over2},\quad -{W\over2}\le y\le{W\over2},\quad -{H\over2}\le z\le{H\over2}. $$ Then we can parameterize the positions of tangency points as follows: $$ P=\left({L\over2},{W\over2}t,{H\over2}u\right),\quad Q=\left({L\over2}u,{W\over2}v,{H\over2}\right),\quad R=\left({L\over2}t,{W\over2},{H\over2}v\right), $$ where parameters $t$, $u$, $v$ vary in the range $-1\le t,u,v\le1$. With this choice the above constraints are automatically fulfilled.

From now on, for the sake of simplicity, I'll consider the case of a cube $L=W=H=2$. The general case can be recovered with a scaling. The above coordinates become: $$ P=\left(1,t,u\right),\quad Q=\left(u,v,1\right),\quad R=\left(t,1,v\right), $$

The generic equation of the ellipsoid is: $$ \alpha x^2+\beta y^2+\gamma z^2+\delta xy+\eta yz+\zeta xz=1. $$ Imposing the ellipsoid to be tangent to the cube at $P$, $Q$, $R$ we can find the coefficients as functions of $t$, $u$, $v$: $$ \alpha={1-v^2\over d},\quad \beta={1-u^2\over d},\quad \gamma={1-t^2\over d},\quad \delta = 2{uv-t\over d},\quad \eta = 2{tu-v\over d},\quad \zeta = 2{tv-u\over d}, $$ where: $$ d = 1 + 2 t u v - t^2 - u^2 - v^2. $$ The matrix associated to the ellipsoid is $$ M=\pmatrix{ \alpha & \delta/2 & \zeta/2 \\ \delta/2 & \beta & \eta/2 \\ \zeta/2 & \eta/2 & \gamma \\ } $$ and we can relate it to the semi-axes $a$, $b$, $c$ of the ellipsoid via the invariants: $$ \det M ={1\over a^2b^2c^2},\quad \text{tr}\ M = {1\over a^2}+{1\over b^2}+{1\over c^2}. $$ Inserting here the above values, we thus find: $$ a^2b^2c^2=d=1 + 2 t u v - t^2 - u^2 - v^2,\quad a^2b^2+b^2c^2+a^2c^2=3-t^2-u^2-v^2, $$ which can be rearranged as: $$ tuv={1\over2}(2-a^2b^2-b^2c^2-a^2c^2+a^2b^2c^2),\quad t^2+u^2+v^2=3-a^2b^2-b^2c^2-a^2c^2. $$ These equalities give then two constraints on the parameters. The third invariant of $M$ would simply give $a^2+b^2+c^2=3$, which is already known.

The second constraint means that point $(t,u,v)$ lies on the surface of a sphere with radius $$ r=\sqrt{3-a^2b^2-b^2c^2-a^2c^2}. $$ We can then set: $$ t=r\cos\theta,\quad u=r\sin\theta\cos\phi,\quad v=r\sin\theta\sin\phi, $$ and inserting into the first constraint we get: $$ \sin^2\theta\cos\theta\sin2\phi= {2-a^2b^2-b^2c^2-a^2c^2+a^2b^2c^2\over r^3}. $$ If we fix the value of $\theta$ we can compute from there the possible values of $\phi$ and parameters $t,u,v$.


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