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I've got the formula for launching to space going Northeast as

$$\cos(inc) = \cos(lat) \sin(az)$$

and that works great but what if I wanted to go southeast?

For example, I know that from a latitude of 28.5 degrees to get to the ISS with it's 51.6 degrees inclination, I need to launch with an azimuth of 44.98 degrees for the northeast path and online tools tell me to use 135.02 degrees for southeast but what is the mathematical formula for that southeast figure?

thanks.

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  • $\begingroup$ If you can add a link to the source of your equation, or at least mention where it comes from, that would be nice. That may help others to formulate the best answer for you. Welcome to Space! $\endgroup$
    – uhoh
    Jun 16, 2019 at 22:52
  • $\begingroup$ Use -51.6 in the same formula? $\endgroup$ Jun 17, 2019 at 0:26

1 Answer 1

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With the space launch complex latitude $\phi$ and the target orbit inclination fixed $i$, you just need to clear out the azimuth $Az$ from your formula as

$Az=\arcsin(\cos(i) \ /\cos(\phi))$,

putting a $\phi$=28.5º and $i$=51.6º, the previous equation has two solutions for the azimuth (remember that $\sin(x)=\sin(\pi-x)$): $Az$=44.98º, 135.02º. So you can launch NorthEast or SouthEast to reach the target inclination. Usually for safety reasons some azimuth are forbiddens.

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  • $\begingroup$ nicely solved!! $\endgroup$
    – uhoh
    Jun 19, 2019 at 11:15

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