Inertia Tensor of Triangle | Inertia Tensor Triangle

The inertia tensor $\mathbf{J}$ of a triangle (like the inertia tensor of any body) can be expressed in terms of covariance $\mathbf{C}$ of the body:

$\mathbf{J} = \mathrm{tr}(\mathbf{C})\mathbf{I} - \mathbf{C}$

where covariance is defined as area integral over the triangle:

$\mathbf{C} \triangleq \int_{\Delta} \rho \mathbf{x}\mathbf{x}^{\mathrm{T}} \, dA$

Covariance for a triangle in three-dimensional space, assuming that mass is equally distributed over the surface with unit density, is

$\mathbf{C} = a \mathbf{V}^{\mathrm{T}} \mathbf{S} \mathbf{V}$

where

represents 3 × 3 matrix containing triangle vertex coordinates in the rows,
is twice the area of the triangle,
$\mathbf{S}= \frac{1}{24} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ \end{bmatrix}$

Substitution of triangle covariance in definition of inertia tensor gives eventually

$\mathbf{J} = \frac{a}{24}(\mathbf{v}^2_0 + \mathbf{v}^2_1 + \mathbf{v}^2_2 + (\mathbf{v}_0 + \mathbf{v}_1 + \mathbf{v}_2)^2)\mathbf{I} - a \mathbf{V}^{\mathrm{T}} \mathbf{S} \mathbf{V}$

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