Inertia Tensor of an Ellipsoidal Shell

2024-08-08 update: the derivation and resulting value of the ellipsoidal shell inertia tensor in the original version of this post was incorrect. The derivation and result have been corrected and the notation has also been updated. A Monte Carlo simulation was also added to verify the result.

In this post we derive the inertia tensor for an ellipsoidal shell with negligible thickness, such that all mass is distributed uniformly across the surface of the ellipsoid. This result is not as easy to find online as the inertia tensors for other common 3D shapes (probably because it is quite ugly). If you're in a hurry, the inertia tensor for an ellipsoidal shell of mass $m$ and semi-axes $a,b,c$ is given in $\eqref{8}$.

Spherical shell inertia

The inertia of a uniform-density spherical shell with mass $m$ and radius $r$ is well-known to be $\bm{I}_{\circ}=(2/3)mr^2\bm{1}_3$, where $\bm{1}_3$ is the $3\times 3$ identity matrix. We use the subscript $(\cdot)_\circ$ to differentiate the inertia of the shell from that of the uniform solid, which we will denote $\bm{I}_\bullet$. All of the inertia tensors in this post are taken about the centroid of the shape, which coincides with the center of mass.

We will derive $\bm{I}_\circ$ for the sphere here before moving on to the more general ellipsoidal case, because the derivations are similar. We will use a similar approach to the derivation here.

Recall that the inertia tensor of a uniform-density solid sphere with mass $m$ and radius $r$ is $\bm{I}_\bullet=(2/5)mr^2\bm{1}_3$. Substituting in $m=\rho V$, where $\rho$ is the density and $V=(4/3)\pi r^3$ is the volume of the sphere, we get $\bm{I}_\bullet=(8/15)\pi\rho r^5\bm{1}_3$.

To obtain a shell, we will take one solid sphere of radius $r$ and subtract its inertia from that of a slightly larger solid sphere, which has radius $r+\delta$ with $\delta\geq 0$, and then taking the limit $\delta\to 0$. This gives us

The density of the shell is given by

where the volume (the denominator) is the difference between the volumes of the two spheres. Substituting $\eqref{2}$ into $\eqref{1}$, we get

We can easily determine that

using the small block of Python code

import sympy
r, δ = sympy.symbols("r, δ")
expr = ((r + δ) ** 5 - r**5) / ((r + δ) ** 3 - r**3)
print(expr.limit(δ, 0))

Substituting $\eqref{4}$ back into $\eqref{3}$, we achieve the expected result for the spherical shell:

Ellipsoidal shell inertia

We are now ready to generalize to the inertia tensor of an ellipsoidal shell, but be warned: the expression is not as nice as that for the sphere!

In this section of the article we'll now use $\bm{I}_\circ$ and $\bm{I}_\bullet$ to refer to the inertia tensors of the ellipsoidal shell and solid ellipsoid, respectively. We will consider an ellipsoid of mass $m$ and semi-axes $a,b,c$. This ellipsoid consists of the set of all points $\bm{x}\in\mathbb{R}^3$ that satisfy $\bm{x}^T\bm{E}^{-1}\bm{x}\leq 1$, where

The inertia tensor of a solid ellipsoid of uniform density is $\bm{I}_\bullet=(1/5)m\bm{S}$, where

with $\mathrm{tr}(\cdot)$ denoting the matrix trace. Again we substitute in $m=\rho V$, where $V=(4/3)\pi abc$ is the volume of the ellipsoid, to obtain $\bm{I}_\bullet=(4/15)\pi\rho abc\bm{S}$.

In the same way as for the spherical shell, we will obtain an ellipsoidal shell by subtracting an ellipsoid with semi-axes $a,b,c$ from a slightly larger ellipsoid with semi-axes $a+\delta,b+\delta,c+\delta$. Taking the limit $\delta\to 0$, we get

with density

and where

Substituting $\eqref{7}$ into $\eqref{6}$, we will again turn to Python to evaluate the limit:

import sympy
δ = sympy.symbols("δ")
a, b, c = sympy.symbols("a, b, c")
S = sympy.Matrix.diag(b**2 + c**2, a**2 + c**2, a**2 + b**2)

a2 = a + δ
b2 = b + δ
c2 = c + δ
S2 = sympy.Matrix.diag(b2**2 + c2**2, a2**2 + c2**2, a2**2 + b2**2)

nom = a2 * b2 * c2 * S2 - a * b * c * S
den = a2 * b2 * c2 - a * b * c
I = (nom / den).limit(δ, 0) / 5
print(I)

Our final result is

where

As I said, not as nice as the spherical shell result! The spherical shell inertia $\eqref{5}$ is of course a special case of the ellipsoidal shell inertia $\eqref{8}$ when $r=a=b=c$, which we can quickly verify with

# continuing from the above code...
r = sympy.symbols("r")
print(I.subs({a: r, b: r, c: r}))

The ellipsoidal shell inertia $\eqref{8}$ is implemented in rigeo.

Monte Carlo Simulation

To verify our result, we can compare our analytical result $\eqref{8}$ with the inertia computed from a large number of point masses uniformly sampled from the surface of an ellipsoid. Using this script, we compare the inertia computed from 100,000 points with equal mass uniformly sampled from an ellipsoid with semi-axes $a=1.0,b=0.75,c=0.5$ to the analytical value we derived above. The sampled value is

the predicted analytical value is

and the difference between them is

which shows that the two values are in good agreement.