Inertia tensor of triangle

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The inertia tensor of a triangle (like the inertia tensor of any body) can be expressed in terms of covariance of the body:

where covariance is defined as area integral over the triangle:

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

where

  • represents 3 × 3 matrix containing triangle vertex coordinates in the rows,
  • is twice the area of the triangle,

Substitution of triangle covariance in definition of inertia tensor gives eventually

A proof of the formula[edit]

The proof given here follows the steps from the article.[1]

Covariance of a canonical triangle[edit]

Let's compute covariance of the right triangle with the vertices (0,0,0), (1,0,0), (0,1,0).

Following the definition of covariance we receive

The rest components of are zero because the triangle is in .

As a result,

Covariance of the triangle with a vertex in the origin[edit]

Consider a linear operator

that maps the canonical triangle in the triangle , , . The first two columns of contain and respectively, while the third column is arbitrary. The target triangle is equal to the triangle in question (in particular their areas are equal), but shifted with its zero vertex in the origin.

Covariance of the triangle in question[edit]

The last thing remaining to be done is to conceive how covariance is changed with the translation of all points on vector .

where

is the centroid of dashed triangle.

It's easy to check now that all coefficients in before is and before is . This can be expressed in matrix form with as above.

References[edit]

  1. ^ http://number-none.com/blow/inertia/bb_inertia.doc Jonathan Blow, Atman J Binstock (2004) "How to find the inertia tensor (or other mass properties) of a 3D solid body represented by a triangle mesh"