620 H Generalized Hammer projection
Following E. Grafarend et al. (1996), we can immediately formulate the series expansion with respect
to an upper triangular matrix R A truncated up to degree five according to (H.123) subject to (H.124)
as a forward substitution.
⎡
⎣
y
y
3
y
5
⎤
⎦ =
⎡
⎣
1 a 13 a 15
0 1 a 35
0 0
1
⎤
⎦
⎡
⎣
x
x
3
x
5
⎤
⎦ + r = R A
⎡
⎣
x
x
3
x
5
⎤
⎦ + r ,
(H.123)
a 13 =
2
3
E
2 , a 15 =
3
5
E
4 , a 35 = 3a 13 = 2E
2 .
(H.124)
In contrast, the backward substitution leads to (H.125) or (H.126).
⎡
⎣
x
x
3
x
5
⎤
⎦ =
⎡
⎣
1 b 13 b 15
0 1 b 35
0 0
1
⎤
⎦
⎡
⎣
y
y
3
y
5
⎤
⎦ + s = R B
⎡
⎣
y
y
3
y
5
⎤
⎦ + s ,
(H.125)
R A R B = I 3 ,
⎡
⎣
1 b 13 b 15
0 1 b 35
0 0
1
⎤
⎦
⎡
⎣
1 a 13 a 15
0 1 a 35
0 0
1
⎤
⎦ =
⎡
⎣
1 0 0
0 1 0
0 0 1
⎤
⎦ .
(H.126)
Finally, we obtain (H.127) and thus (H.128).
b 13 = −a 13 , b 15 = 3a
2
13 − a 15 , b 35 = −3a 13 ,
(H.127)
b 13 = −
2
3
E
2 , b 15 =
11
15
E
4 .
(H.128)
Using the first row of R B , we arrive at (H.129).
x = y + b 13 y
3 + b 15 y
5 + O(y
7 ) .
(H.129)
The powers of y are computed according to (H.130), leading to (H.131).
y = c 4
sin Φ +
2
3
E
2 sin
3 Φ +
3
5
E
4 sin
5 Φ + O(E
6 )
,
y
3 = c
3
4
sin
3 Φ + 2E
2 sin
5 Φ +
47
15
E
4 sin
7 Φ + O(E
6 )
,
y
5 = c
5
4
sin
5 Φ +
10
3
E
2 sin
7 Φ +
67
9
E
4 sin
9 Φ + O(E
6 )
,
(H.130)
sin Φ
∗ = c 4 sin Φ
1 +
2
3
E
2 sin
2 Φ(1 − c
2
4 ) +
1
15
E
4 sin
4 Φ(9 − 20c
2
4 + 11c
4
4 ) + O(E
6 )
. (H.131)