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Old 05-25-2019, 09:10 AM   #1
Nikogori
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Default Derivation of launch azimuth equation



There are several different approaches to derive this formula. I'm going to show you one example using basic spherical trigonometry because I believe it is the easiest method. (and I need someone to check if I'm doing it right)



Yellow is target orbit. Red is equator. Blue is meridian passing through the launch site.
Launch azimuth is the angle between Yellow and Blue.
Angle between red and blue is 90.

Arc length of a side of a spherical triangle can be expressed in angle.
For example, arc length from equator to launch site can be expressed in latitude.



b is launch site latitude. (b = ∠AOC)
B is target inclination.
A is launch azimuth.
also, C = 90

Now we're thinking with unit sphere. Distance from the center to the surface is always 1.

Draw a perpendicular line from point A to OB. It intersects OB at M.
AM = sin c

Draw a perpendicular line from point A to OC. It intersects OC at L.
AL = sin b

∠AML = B
AM*sin B = AL
sin c sin B = sin b

Same logic can be applied to A and a.
sin c sin A = sin a (1)
(spherical sine rule when C = 90)

Draw a line between L and M.
∠OML = 90

AM*cos B = LM
sin c cos B = LM

OL = cos b
OL*sin a = LM
cos b sin a = LM

Therefore,
sin c cos B = cos b sin a (2)
sin c cos A = cos a sin b

from (1) and (2),
sin A = cos B / cos b
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