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Hi guys,

I am doing navigation mathematics and I met all kinds of references regarding how the attitude update equation should be solved. However, none of the references that I found makes any sense to me.

I wonder if someone could enlighten me with this math.

The attitude update equation for direction cosine matrix is given by:

[tex]

\dot{\mathbf{C}}_b^i = \mathbf{C}_b^i \mathbf{\Omega}_{ib}^b

[/tex]

where [tex]\mathbf{C}_b^i[/tex] indicates the transformation matrix from the b-frame to the i-frame and [tex]\mathbf{\Omega}_{ib}^b[/tex] indicates the skew-symmetric angular velocity matrix of the b-frame with respect to the i-frame resolved in the b-frame.

I really wonder if there exists some kind of closed-form solution for this equation. I do think that there should be one way or to express the exact solution to this matrix differential equation because if we look at the transpose:

[tex]

\dot{\mathbf{C}}_n^b = -\mathbf{\Omega}_{nb}^b \mathbf{C}_n^b

[/tex]

by using the orthogonality of the direction cosine and skew-symmetric properties.

But this last equation is very similar to the state transition matrix equation:

[tex]

\mathbf{\dot{\Phi}}=\mathbf{A}\mathbf{\Phi}

[/tex]

which has the solution of a Peano-Baker series:

[tex]

\mathbf{\Phi} =& \mathbf{I} + \int_{t_o}^t \mathbf{A}(\sigma_1)d\sigma_1 + \int_{t_0}^t \mathbf{A}(\sigma_1) \int_{t_0}^{\sigma_1} \mathbf{A}(\sigma_2) d\sigma_2 d\sigma_1 \\

& + \int_{t_0}^t \mathbf{A}(\sigma_1) \int_{t_0}^{\sigma_1} \mathbf{A}(\sigma_2) \int_{t_0}^{\sigma_2} \mathbf{A}(\sigma_3) d\sigma_3d\sigma_2 d\sigma_1 + \cdots

[/tex]

Worse still, most of the time the navigation is done on the local navigation frame in which we have to transform the earth's rate and local transport rate into the equation. Hence, the differential equation looks like

[tex]

\dot{\mathbf{C}}_b^n = \mathbf{C}_b^n \mathbf{\Omega}_{ib}^b - \left(\mathbf{\Omega}_{ie}^n + \mathbf{\Omega}_{ne}^n\right)\mathbf{C}_b^n

[/tex]

So, I am thinking that the solution should be along that line... but I am not so sure...

I would greatly appreciate any help. :)

I am doing navigation mathematics and I met all kinds of references regarding how the attitude update equation should be solved. However, none of the references that I found makes any sense to me.

I wonder if someone could enlighten me with this math.

The attitude update equation for direction cosine matrix is given by:

[tex]

\dot{\mathbf{C}}_b^i = \mathbf{C}_b^i \mathbf{\Omega}_{ib}^b

[/tex]

where [tex]\mathbf{C}_b^i[/tex] indicates the transformation matrix from the b-frame to the i-frame and [tex]\mathbf{\Omega}_{ib}^b[/tex] indicates the skew-symmetric angular velocity matrix of the b-frame with respect to the i-frame resolved in the b-frame.

I really wonder if there exists some kind of closed-form solution for this equation. I do think that there should be one way or to express the exact solution to this matrix differential equation because if we look at the transpose:

[tex]

\dot{\mathbf{C}}_n^b = -\mathbf{\Omega}_{nb}^b \mathbf{C}_n^b

[/tex]

by using the orthogonality of the direction cosine and skew-symmetric properties.

But this last equation is very similar to the state transition matrix equation:

[tex]

\mathbf{\dot{\Phi}}=\mathbf{A}\mathbf{\Phi}

[/tex]

which has the solution of a Peano-Baker series:

[tex]

\mathbf{\Phi} =& \mathbf{I} + \int_{t_o}^t \mathbf{A}(\sigma_1)d\sigma_1 + \int_{t_0}^t \mathbf{A}(\sigma_1) \int_{t_0}^{\sigma_1} \mathbf{A}(\sigma_2) d\sigma_2 d\sigma_1 \\

& + \int_{t_0}^t \mathbf{A}(\sigma_1) \int_{t_0}^{\sigma_1} \mathbf{A}(\sigma_2) \int_{t_0}^{\sigma_2} \mathbf{A}(\sigma_3) d\sigma_3d\sigma_2 d\sigma_1 + \cdots

[/tex]

Worse still, most of the time the navigation is done on the local navigation frame in which we have to transform the earth's rate and local transport rate into the equation. Hence, the differential equation looks like

[tex]

\dot{\mathbf{C}}_b^n = \mathbf{C}_b^n \mathbf{\Omega}_{ib}^b - \left(\mathbf{\Omega}_{ie}^n + \mathbf{\Omega}_{ne}^n\right)\mathbf{C}_b^n

[/tex]

So, I am thinking that the solution should be along that line... but I am not so sure...

I would greatly appreciate any help. :)

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