Talk:Launch Azimuth

about rotation compensation: wouldn't it be better if that triangle was solved using cosine and sine theorems? Both explanation and example then could be shorter

Yes, these theorems make sense, but I think not all people in the Orbiter community will know what you mean with them.--Urwumpe 14:36, 3 November 2007 (MSK)

Modified Equatorial Radius

In the Rotation of the Earth Section, the author used the mean radius of Earth. The variable is r(eq) which is 6378.137km according to the WGS84 ellipsoid. http://en.wikipedia.org/wiki/Earth_radius

Launch direction confusion

In the example for the relation between inclination and azimuth, the article states that: "As a launch on the southeastern course would overfly the Bahamas, Cuba, and South America, the northeastern course is always used. This is ocean all the way to Ireland." I would have expected that the northeastern course is always because if you go the other way, you will end up in orbit - going the opposite direction to the ISS!?

No, you have two directions for getting into an orbit, if the inclination of the orbit is bigger than your latitude. Remember, the ground track of the orbit is essentially a sine wave. It goes up to the north and then to the south again. You can launch into this orbit, when the ground track passes over your launch site. Even if you have to launch eastward to enter this orbit, you can still launch at a time, when the ground track goes to the northeast, or when it goes to the southeast - both is eastern direction. The opposite direction to the ISS would be northwest or southwest. --Urwumpe (talk) 09:24, 13 December 2013 (UTC)

Rotation of the Earth

For homogeneity, wouldn't it be better to express the speed in m/s instead of km/s here ?

"The third is just the difference between those two vectors. Note that you have to know the speed of your target orbit! Calculate this from the orbit altitude, or use 7.730km/s for a typical 300km circular orbit.".