PlanetPhysics/Euler's Moment Equations

Euler's Moment Equations in terms of the principle axes is given by

$M_{x}=I_{x}{\dot {\omega _{x}}}+(I_{z}-I_{y})\omega _{y}\omega _{z}$ $M_{y}=I_{y}{\dot {\omega _{y}}}+(I_{x}-I_{z})\omega _{x}\omega _{z}$ $M_{z}=I_{z}{\dot {\omega _{z}}}+(I_{y}-I_{x})\omega _{x}\omega _{y}$

In order to derive these equations, we start with the angular momentum of a rigid body

${\vec {H_{B}}}=I\omega =\left[{\begin{matrix}{c c c}I_{xx}&-I_{xy}&-I_{xz}\\-I_{yx}&I_{yy}&-I_{yz}\\-I_{zx}&-I_{zy}&I_{zz}\end{matrix}}\right]\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]$

Since the vector is in the body frame and we want the Moment in an inertial frame we need to use the transport theorem since our body is in a non-inertial reference frame to express the derivative of the angular momentum vector in this frame. So the Moment is given by

${\vec {M}}={\dot {\vec {H_{I}}}}={\dot {\vec {H_{B}}}}+{\vec {\omega }}\times {\vec {H_{B}}}$

Since we are assuming the inertia tensor is expressed using the principal axes of the body the Products of Inertia are zero

$I_{yx}=I_{xy}=I_{xz}=I_{zx}=I_{zy}=I_{yz}=0$

and using the shorter notation

$I_{xx}=I_{x}$ $I_{yy}=I_{y}$ $I_{zz}=I_{z}$

Also since the moments of inertia are constant, when we take the derivative of the Inertia Tenser it is zero, so

${\dot {\vec {H_{B}}}}=\left[{\begin{matrix}{c c c}I_{x}&0&0\\0&I_{y}&0\\0&0&I_{z}\end{matrix}}\right]\left[{\begin{matrix}{\dot {\omega _{x}}}\\{\dot {\omega _{y}}}\\{\dot {\omega _{z}}}\end{matrix}}\right]+\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]\times \left[{\begin{matrix}{c c c}I_{x}&0&0\\0&I_{y}&0\\0&0&I_{z}\end{matrix}}\right]\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]$

Carrying out the matrix multiplication ${\dot {\vec {H_{B}}}}=\left[{\begin{matrix}I_{x}{\dot {\omega _{x}}}\\I_{y}{\dot {\omega _{y}}}\\I_{z}{\dot {\omega _{z}}}\end{matrix}}\right]+\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]\times \left[{\begin{matrix}I_{x}\omega _{x}\\I_{y}\omega _{y}\\I_{z}\omega _{z}\end{matrix}}\right]$

after evaluating the cross product, we are left with adding the vectors ${\dot {\vec {H_{B}}}}=\left[{\begin{matrix}I_{x}{\dot {\omega _{x}}}\\I_{y}{\dot {\omega _{y}}}\\I_{z}{\dot {\omega _{z}}}\end{matrix}}\right]+\left[{\begin{matrix}\omega _{y}\omega _{z}(I_{z}-I_{y})\\\omega _{x}\omega _{z}(I_{x}-I_{z})\\\omega _{x}\omega _{y}(I_{y}-I_{x})\end{matrix}}\right]$

Once we add these vectors we are left with Euler's Moment Equations

$M_{x}=I_{x}{\dot {\omega _{x}}}+(I_{z}-I_{y})\omega _{y}\omega _{z}$ $M_{y}=I_{y}{\dot {\omega _{y}}}+(I_{x}-I_{z})\omega _{x}\omega _{z}$ $M_{z}=I_{z}{\dot {\omega _{z}}}+(I_{y}-I_{x})\omega _{x}\omega _{y}$