1
$\begingroup$

I propagate the attitude of a satellite in a free-floating, torque free environment from $t_k$ to $t_{k+1}$ by integrating Euler's equation of motion:

$$ \dot{\omega} = [J]^{-1}([\omega]_{\times}[J]\:\omega)$$

with $[J]\in\mathbb{R}^{3\times3}$ being the (full) inertia matrix and $\omega \in \mathbb{R}^3$ the angular velocity vector, and $[\cdot]_{\times}$ being the skew-symmetric operator.

If we consider an Euler integration step of $dt$, we can therefore write:

$$ \omega_{k+1} = \dot{\omega}_k\cdot dt = [J]^{-1}([\omega_k]_{\times}[J]\:\omega_k)\cdot dt$$

I would like to compute the analytical Jacobian $\textbf{F}_\omega = \frac{\partial\omega_{k+1}}{\partial \omega_k}$, or an approximation of it, in order to propagate the covariance from $t_k$ to $t_{k+1}$. Is there any known closed-form solution? With the cross-product, I really don't know how to tackle the problem.

Any help is welcome. Thank you.

$\endgroup$
2
  • $\begingroup$ Based on the comment you make near the end of your question: does it help that the derivative of (a cross b) is a’ cross b + a cross b’ ? $\endgroup$
    – SteveO
    May 25 at 17:21
  • $\begingroup$ It does, thank you very much! $\endgroup$
    – Maltergate
    May 26 at 8:39
1
$\begingroup$

$$\frac{\partial\dot\omega}{\partial\omega} = [J]^{-1}\frac{\partial([\omega]_{\times}[J]\omega)}{\partial\omega}$$

$$= [J]^{-1} \frac{\partial(\omega\times[J]\omega)}{\partial\omega}$$

$$= [J]^{-1} \left (\frac{\partial\omega}{\partial\omega}\times[J]\omega + \omega \times [J]\frac{\partial\omega}{\partial \omega}\right)$$

$$ = [J]^{-1} \left (1\times[J]\omega + [\omega]_\times[J] \right) $$

Using the two definitions:

  1. $a \times b = [a]_{\times}b = [b]^{\intercal}_{\times}a$
  2. $[x]^{\intercal}_{\times} = -[x]_{\times}$

on the term $1\times[J]\omega$, we find the correct answer: $$ \frac{\partial\dot\omega}{\partial\omega} = [J]^{-1} \left ( [\omega]_{\times}[J] - \left [[J]\omega \right]_{\times} \right )$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.