Take the 2-minute tour ×
Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's 100% free, no registration required.

I've been looking into the document IS-GPS-200H to understand how to calculate satellite location in the ECEF coordinate.

I am having problem understanding the formula to derive $\Omega$, the longitude of the ascending node (LAN) relative to Greenwich at given time $t$:

$$ \Omega = \Omega_0 + \left( \dot{\Omega - w} \right)\times t_k - w \times t_{oe} $$

where: $$ \Omega_0: \text{LAN relative to vernal equinox, at the beginning of the week}\\ \dot{\Omega}: \text{angular velocity for LAN, relative to vernal equinox.}\\ w: \text{angular velocity of earth, relative to vernal equinox.}\\ t_k: t - t_{oe}\\ t_{oe}: \text{ephemeris reference epoch}\\ $$ (and let us denote the beginning of the week as $t_0$ for brevity).

But if I try to work out this from scratch:

  1. At $t = t_0$, LAN was $\Omega_0$. But since what we really need is the difference of LAN and longitude of Greenwich, we also need to know $w_0$, the initial longitude of Greenwich at $t = t_0$. $$ \Omega(t = t_0) = \Omega_0 - w_0 $$
  2. At the ephemeris reference epoch time $t = t_{oe}$, LAN and the earth both rotate with their respective angular momentum and hence: $$ \Omega(t = toe) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe} $$
  3. As time varies from $toe$ to $t$, again LAN and the earth both rotate with their own respective angular momentum and hence $$ \Omega(t) = \Omega_0 + w_0 + (\dot{\Omega} - w) \times t_{oe} + (\dot{\Omega} - w) \times t_k $$ which obviously differs from the right formula by $w_0 + \dot{\Omega} \times t_{oe}$.

My question is where am I making mistakes/misunderstanding the eqution? Explain also why we don't need to know $w_0$ or equivalent input, that would be greatly appreciated.

share|improve this question
    
I just edited your questions check out this guide which explains how you can enter equations for future reference. –  harogaston Jul 13 at 4:47

1 Answer 1

I think you're not too far from understanding the formula. However, there are a couple details you're missing:

  1. From what I understand from the document you linked to, $t_0 = t_{oe}$ is the time at the beginning of the week, or reference epoch.
  2. When computing an angle from a angular velocity, you usually need to compute it with respect to an initial angle and an initial time: when the Earth and the satellite rotate between $t_0$ and $t$, the angle they rotated of is $\dot{\Omega}(t - t_0)$ or $\dot{\Omega}_E(t - t_0)$.
    The only exception to this rule seems to be the initial Greenwich longitude with respect to the vernal equinox, which seems to be computed from $t = 0$.
  3. $w_0$, the initial Greenwich longitude wrt the vernal equinox, is $\dot{\Omega}_E t_{oe}$. From this formula, it seems like the document considers an origin of time $t = 0$ when the longitude of Greenwich wrt to the vernal equinox at this time is 0.

To summarize: the formula would have been more explicit and comprehensible written as: $$ \Omega(t) = \Omega_0 - \dot{\Omega}_E t_{oe} + (\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe}) $$ with:

  • $\Omega_0$ = LAN wrt the vernal equinox at $t_{oe}$
  • $\dot{\Omega_E}$ = angular velocity of the Earth wrt the vernal equinox
  • $-\dot{\Omega}_Et_{oe}$ = Longitude of Greenwich at $t_{oe}$
  • $\dot{\Omega}$ = angular velocity of the satellite wrt the vernal equinox
  • $(\dot{\Omega} - \dot{\Omega}_E)(t - t_{oe})$ = relative angular displacement between the satellite and Greenwich, between time $t$ and $t_{oe}$

That said, it seems like really bad practice to name $\Omega(t)$ the longitude with respect to Greenwich and $\dot{\Omega}$ the angular velocity with respect to the vernal equinox. This causes quite a bit of confusion, since mathematically speaking, for any linear function $f$, $f(t) = f(t_0) + \dot{f}\times(t - t_0)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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