Timeline for Conversion of velocity vector in ITRF to GCRF
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2022 at 14:02 | comment | added | David Hammen | @GuillaumeJ The + (as in your example) is used when going from body-fixed to inertial. The expression I used converts from inertial (GCRF) to body-fixed (ITRF), so here subtraction is needed rather than addition. | |
| Mar 16, 2022 at 13:59 | history | edited | David Hammen | CC BY-SA 4.0 |
ITRF, not ICRF
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| Mar 16, 2022 at 8:55 | comment | added | GuillaumeJ | @DavidHammen And why in the expression you gave above it is minus $\vec{w}$...? In every sources of the transport theorem that I found, it's always "+" ; here for example. | |
| Mar 15, 2022 at 15:29 | comment | added | GuillaumeJ | @DavidHammen Thanks a lot! $\vec{w}$ is just $w_x=0, w_y=0, w_z=7.2921159 × 10^{−5}$ ? Or is it more complicated ? If it is, can you give me any documentation on how to compute this vector ? | |
| Mar 15, 2022 at 14:49 | comment | added | David Hammen | @GuillaumeJ Zero, or nearly so. That is not a good way to calculate the derivative of a transformation matrix. A much better approach is to use $\dot{\mathbf{R}}_{R\to I} = \mathbf{R}_{R\to I} \mathbf{Sk}(\vec \omega)$ where $\mathbf{R}_{R\to I}$ is the transformation matrix from Earth-fixed to Earth-centered inertial, $\mathbf{Sk}(\vec \omega)$ is the skew-symmetric cross product matrix generated from $\vec\omega$, and $\vec\omega$ is the angular velocity of the Earth with respect to inertial, expressed in Earth-fixed coordinates. | |
| Mar 15, 2022 at 14:39 | vote | accept | Rafa | ||
| Mar 15, 2022 at 14:11 | comment | added | GuillaumeJ | @DavidHammen By taking the time derivative of the RGCRF→ICRF matrix, what would be the accurate deltaT to do this computation (R' = (R(t+deltaT)-R(t))/deltaT) ? | |
| Feb 3, 2022 at 1:44 | comment | added | Rafa | Thanks a lot, this clears up many things! So it seems the key is how accurate we want to be when calculating $\omega$. I have seen that Earth orientation parameters give the exact duration of a day (in seconds of deviation from the standard 86400s). For the purposes of propagation of orbits of artificial satellites, I think I will try the approximate way. Is there any applications where the exact way (getting the messy time derivative you mentioned) would be required? | |
| Feb 3, 2022 at 1:27 | history | edited | Rafa | CC BY-SA 4.0 |
added missing curly bracket in mathjax
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| Feb 2, 2022 at 17:07 | history | edited | David Hammen | CC BY-SA 4.0 |
added 27 characters in body
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| Feb 2, 2022 at 16:15 | comment | added | David Hammen | Is MathJax broken today, or just slow? | |
| Feb 2, 2022 at 16:15 | history | answered | David Hammen | CC BY-SA 4.0 |