Difference between revisions of "Talk:Launch Azimuth"

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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
 
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
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== Launch direction confusion ==
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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!?

Revision as of 12:00, 12 December 2013

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!?