Editing Moment of inertia (section)

=== Ellipsoid ===
[[File:Triaxial Ellipsoid.jpg|thumb|right|An ellipsoid with the semi-principal diameters labelled <math>a</math>, <math>b</math>, and <math>c</math>.]]
The moment of inertia matrix in body-frame coordinates is a quadratic form that defines a surface in the body called [[Poinsot's ellipsoid]].<ref>
{{cite book
 |first=Matthew T. |last=Mason
 |url=https://books.google.com/books?id=Ngdeu3go014C
 |title=Mechanics of Robotics Manipulation
 |publisher=MIT Press
 |year=2001
 |isbn=978-0-262-13396-8
 |access-date=November 21, 2014
}}</ref> Let <math>\boldsymbol{\Lambda}</math> be the inertia matrix relative to the center of mass aligned with the principal axes, then the surface
<math display="block">\mathbf{x}^\mathsf{T}\boldsymbol{\Lambda}\mathbf{x} = 1,</math>
or
<math display="block">I_1x^2 + I_2y^2 + I_3z^2 =1,</math>
defines an [[ellipsoid]] in the body frame. Write this equation in the form,
<math display="block"> \left(\frac{x}{1/\sqrt{I_1}}\right)^2 + \left(\frac{y}{1/\sqrt{I_2}}\right)^2 + \left(\frac{z}{1/\sqrt{I_3}}\right)^2 = 1,</math>
to see that the semi-principal diameters of this ellipsoid are given by
<math display="block">a = \frac{1}{\sqrt{I_1}}, \quad b=\frac{1}{\sqrt{I_2}}, \quad c=\frac{1}{\sqrt{I_3}}.</math>

Let a point <math>\mathbf{x}</math> on this ellipsoid be defined in terms of its magnitude and direction, <math>\mathbf{x} = \|\mathbf{x}\|\mathbf{n}</math>, where <math>\mathbf{n}</math> is a unit vector. Then the relationship presented above, between the inertia matrix and the scalar moment of inertia <math>I_\mathbf{n}</math> around an axis in the direction <math>\mathbf{n}</math>, yields
<math display="block">\mathbf{x}^\mathsf{T}\boldsymbol{\Lambda}\mathbf{x} = \|\mathbf{x}\|^2\mathbf{n}^\mathsf{T}\boldsymbol{\Lambda}\mathbf{n} = \|\mathbf{x}\|^2 I_\mathbf{n} = 1. </math>

Thus, the magnitude of a point <math>\mathbf{x}</math> in the direction <math>\mathbf{n}</math> on the inertia ellipsoid is
<math display="block"> \|\mathbf{x}\| = \frac{1}{\sqrt{I_\mathbf{n}}}.</math>
<!---duplicated above
=== Parallel axis theorem for the inertia matrix ===
{{Main|Parallel axis theorem}}
It is useful to note here that if the moment of inertia matrix or tensor is relative to the [[center of mass]], then it can be determined relative to any other reference point in the body using the parallel axis theorem.  If [I{{sub|C}}{{sup|B}}] is the moment of inertia matrix in the body frame relative to the center of mass '''C''', then the moment of inertia matrix [I{{sub|R}}{{sup|B}}] in the same frame but relative to a different point '''R''' is given by
<math display="block">[I_R^B] = [I_C^B] - M[d]^2, </math>
where M is the mass of the body, and [d] is the skew-symmetric matrix obtained from the vector '''d''' = '''R'''&nbsp;−&nbsp;'''C'''.

The tensor form of the parallel axis theorem is given by
<math display="block"> \mathbf{I}_R^B = \mathbf{I}_C^B + M((\mathbf{d} \cdot \mathbf{d}) \mathbf{E} - \mathbf{d} \otimes \mathbf{d}).
</math>
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