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9
=
0
@
1
A
e
2
2
þ
e
4
8
þ
e
6
16
e
2
16
k
0
2
3
128
e
2
k
0
4
0
e
2
ʱ
ﾼ
1
þ
þ
2
4
3
5
e
2
16
k
0
2
e
2
32
k
0
4
0
e
2
ʲ
ﾼ ˁ
1
þ
,
;
e
2
256
k
0
4
0
ʳ
ﾼ ˁ
k
0
2
e
2
cos
2
m
ﾼ
ʳ
0
is 0.0002
00
; hence, the
ʳ
0
term in (
5.100
) is generally
The maximum value of
negligible.
For the approximate solutions in meters,
' can be calculated according
to (
5.92
); for the approximate solutions in hectometers, one can calculate
according to equations that leave out the term of k'
2
:
ʱ
' and
ʲ
e
2
e
2
8
1
2
þ
0
ʱ
ﾼ
:
Finally, one obtains:
L
2
ﾼ
L
1
þ
l
ð
5
:
101
Þ
Determination of the Quadrant
With the above formulae, m, M,
ʻ
2
, and A
2
are calculated using trigonometric
functions. Therefore, it is still necessary to discuss the determination of their
quadrants.
To determine quadrants for these quantities easily, we draw Figs.
5.44
,
5.45
,
5.46
, and
5.47
, which denote the quadrants in which m, M,
ʻ
1
,
ʻ
2
, and A
2
lie when
point P
1
0
is in the northern hemisphere, i.e., u
1
is positive, and A
1
lies in quadrants I,
II, III, and IV. In each figure, P
2
is assumed to have three positions, denoted by P
0
2
,
P
0
2
, and P
00
2
, respectively. Correspondingly,
ʻ
1
,
ʻ
2
lies in three different quadrants. In
the figures, the great circle intersects the equator at P
0
0
and P
0
00
, and dashed lines
denote the back side of the sphere. Figure
5.44
demonstrates that A
1
is in quadrant I,
and here m, M, and
ʻ
1
are all in quadrant I. At points P
0
2
,
ʻ
2
and A
2
0
are both in the
quadrant I; at points P
0
2
,
ʻ
2
and A
0
2
are both in quadrant II; at points P
00
2
,
ʻ
2
lies in
quadrant III while A
2
0
180
, A
2
lies in
quadrant III or quadrant IV. Similarly, we can also explain the situations in
Figs.
5.45
,
5.46
, and
5.47
.
Fig.
5.44
shows that
A
0
2
lies in quadrant II. Since A
2
ﾼ
) are in the same quadrant. The quadrant of A
2
can be determined by the sign of tan A
2
: when tan A
2
is positive, A
2
lies in quadrant
ʻ
2
and (M +
˃
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