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Equation (iii) :
cos
ω
=0
,
sin
ω
=1
ω
=90
◦
⇒
(3.97)
Equation (iv) :
Ω
= 270
◦
+
λ
0
,λ
0
=90
◦
+
Ω.
cos
Ω
=sin
λ
0
,
sin
Ω
=
−
cos
λ
0
⇒
(3.98)
Equation (v) :
I
= 270
◦
+
φ
0
,φ
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
,φ
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
,φ
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
,φ
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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