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Equation (iii) :
cos ω =0 , sin ω =1
ω =90
(3.97)
Equation (iv) :
Ω = 270 + λ 0 0 =90 + Ω.
cos Ω =sin λ 0 , sin Ω =
cos λ 0
(3.98)
Equation (v) :
I = 270 + φ 0 0 =90 + I.
cos I =sin φ 0 , sin I =
cos φ 0
(3.99)
Lemma 3.4 (The transformation of the first oblique frame of reference to the second oblique
frame of reference: λ 0 0
ω,I,Ω ).
If the first oblique frame of reference is given by defining a meta-North Pole
, then the
second oblique frame of reference is determined by the orbital Kepler elements ω =90 ,I =
270 + φ 0 ,and Ω = 270 + λ 0 .
{
λ 0 0
}
End of Lemma.
Lemma 3.5 (The transformation of the second oblique frame of reference to the first oblique
frame of reference: ω,I,Ω
λ 0 0 ).
If the second oblique frame of reference is given by defining the orbital Kepler elements
,
then the first oblique frame of reference is determined by the meta-North Pole λ 0 =90 + Ω and
φ 0 =90 + I , subject to ω =90 .
{
ω,I,Ω
}
End of Lemma.
For the transverse frame of reference, the inclination of the ascending node I is chosen ninety
degrees, i.e. I =90 . Accordingly, the transformation of reference frames leads us to Corollary 3.6 .
Corollary 3.6 (Transformation of reference frames, transverse aspect, I =90 ).
If the second transverse frame of reference is given by defining the orbital Kepler elements as
{ω,I,Ω} = { 90 , 90 ,Ω} , then the first transverse frame of reference is determined by the meta-
North Pole λ 0 =90 + Ω and φ 0 =0 .
End of Corollary.
3-36 Numerical Examples
The following Table 3.1 and the following Table 3.2 show some selected examples of the two
designs to be considered here.
 
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