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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:
![{\displaystyle \mathbf {J} =\mathrm {tr} (\mathbf {C} )\mathbf {I} -\mathbf {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/93f4d60113e30c2fc1422a336bcd7540c3021c4c)
where covariance is defined as area integral over the triangle:
![{\displaystyle \mathbf {C} \triangleq \int _{\Delta }\rho \mathbf {x} \mathbf {x} ^{\mathrm {T} }\,dA}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f17cf6d5b7cf7504f532a3003ae95623035cdbc)
Covariance for a triangle in three-dimensional space, assuming that mass is equally distributed over the surface with unit density, is
![{\displaystyle \mathbf {C} =a\mathbf {V} ^{\mathrm {T} }\mathbf {S} \mathbf {V} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d30cd7caf806506f478f14723e565d2f167e1b61)
where
Substitution of triangle covariance in definition of inertia tensor gives eventually
![{\displaystyle \mathbf {J} ={\frac {a}{24}}(\mathbf {v} _{0}^{2}+\mathbf {v} _{1}^{2}+\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} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fe5303c54963bd3a923e6cc0c6891ef654d467e)
A proof of the formula
The proof given here follows the steps from the article.[1]
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
![{\displaystyle \mathbf {C} _{xx}^{0}=\int _{\Delta }x^{2}\,dA=\int _{x=0}^{1}x^{2}\int _{y=0}^{1-x}\,dy\,dx=\int _{0}^{1}x^{2}(1-x)\,dx={\frac {1}{12}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5565514fa36edeec582177301c81b3d740f6fe5a)
![{\displaystyle \mathbf {C} _{xy}^{0}=\int _{\Delta }xy\,dA=\int _{x=0}^{1}x\int _{y=0}^{1-x}y\,dy\,dx=\int _{0}^{1}x{\frac {(1-x)^{2}}{2}}\,dx={\frac {1}{24}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/236659d31d33e0d35be3f555d1ed1e5d949e788a)
![{\displaystyle \mathbf {C} _{yy}^{0}=\mathbf {C} _{xx}^{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c7677fd9897296e06ebb01f5472cccbe5c4516a)
The rest components of
are zero because the triangle is in
.
As a result
![{\displaystyle \mathbf {C} ^{0}={\frac {1}{24}}{\begin{bmatrix}2&1&0\\1&2&0\\0&0&0\\\end{bmatrix}}={\frac {1}{48}}{\begin{bmatrix}1\\-1\\0\end{bmatrix}}{\begin{bmatrix}1&-1&0\end{bmatrix}}^{\mathrm {T} }+{\frac {1}{16}}{\begin{bmatrix}1\\1\\0\end{bmatrix}}{\begin{bmatrix}1&1&0\end{bmatrix}}^{\mathrm {T} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42962892a2c60544237b8cbfa31686849eb199f6)
Covariance of the triangle with a vertex in the origin
Consider a linear operator
![{\displaystyle \mathbf {x} '=\mathbf {A} \mathbf {x} ^{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b981f0f75538890b933b2eb98e8b0fd91ee17d6a)
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.
![{\displaystyle \mathbf {C} '=\int _{\Delta '}\mathbf {x} '\mathbf {x} '^{\mathrm {T} }\,dA'=\int _{\Delta ^{0}}\mathbf {A} \mathbf {x} ^{0}\mathbf {x} ^{0\mathrm {T} }\mathbf {A} ^{\mathrm {T} }a\,dA^{0}=a\mathbf {A} \mathbf {C} ^{0}\mathbf {A} ^{\mathrm {T} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2360762c06209be5842ba6a10e61012b1137acf9)
![{\displaystyle \mathbf {C} '={\frac {a}{48}}(\mathbf {v} _{1}-\mathbf {v} _{2})(\mathbf {v} _{1}-\mathbf {v} _{2})^{\mathrm {T} }+{\frac {a}{16}}(\mathbf {v} _{1}+\mathbf {v} _{2}-2\mathbf {v} _{0})(\mathbf {v} _{1}+\mathbf {v} _{2}-2\mathbf {v} _{0})^{\mathrm {T} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65c6df205d3494a11375124e931f3c01f766bad1)
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
.
![{\displaystyle \mathbf {C} =\int _{\Delta }(\mathbf {x'} +\mathbf {v} _{0})(\mathbf {x'} +\mathbf {v} _{0})^{\mathrm {T} }\,dA=\mathbf {C} '+{\frac {a}{2}}(\mathbf {v} _{0}\mathbf {v} _{0}^{\mathrm {T} }+\mathbf {v} _{0}{\overline {\mathbf {x} }}'^{\mathrm {T} }+{\overline {\mathbf {x} }}'\mathbf {v} _{0}^{\mathrm {T} })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eaefab45ea69666c1b66dec90b46a34f21a191c0)
where
![{\displaystyle {\overline {\mathbf {x} }}'=\int _{\Delta '}\mathbf {x} '\,dA'={\frac {1}{3}}(\mathbf {v} '_{1}+\mathbf {v} '_{2})={\frac {1}{3}}(\mathbf {v} _{1}+\mathbf {v} _{2}-2\mathbf {v} _{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb2bf03fbcc2057c19a58fb5cbc898c47341ddaf)
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
43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.