The inertia tensor  of a triangle

Triangle

A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices A, B, and C is denoted ....

(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
  
  

  Covariance of a canonical triangle

  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

  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

  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.
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