466 21 Datum problems
Box 21.8 (Decomposition of the synthesis matrix B = B 0 + hB 1 + O 2 (h
2 ). ).
B =
=
2
4
sin l
(n+h) cos b
−
cos l
(n+h) cos b
0
−
[(1−e
2 )n+h] sin b cos l
(n+h) cos b
−
[(1−e
2 )n+h] sin b sin l
(n+h) cos b
sin b cos l
m+h
sin b sin l
m+h
−
cos b
m+h
[n
2 (1−e
2 )+mh] sin l
m(m+h)
−
[n
2 (1−e
2 )+mh] cos l
m(m+h)
· · ·
· · ·
1
0
0
0
0
e
2 n sin b cos b
m+h
−
e
2 n sin b cos b
a 1 (m+h)
−
[e
2 m sin
2 b+2(1−e
2 )n] sin b cos b
2(m+h)(1−e 2 )
3
5 ∈ R
2×9 ,
(21.43)
B 0 :=
=
" sin l
n cos b
−
cos l
n cos b
0
−(1 − e
2 ) tan b cos l −(1 − e
2 ) tan b sin l
sin b cos l
m
sin b sin l
m
−
cos b
m
[n
2 (1−e
2 )] sin l
m 2
−
[n
2 (1−e
2 )] cos l
m 2
· · ·
· · ·
1
0
0
0
0
e
2 n sin b cos b
m
−
e
2 n sin b cos b
a 1 m
−
e
2 m sin
3 b cos b+2(1−e
2 )n sin b cos b
2m(1−e 2 )
3
5 ,
(21.44)
B 1 :=
=
2
4
−
sin l
n 2 cos b
cos l
n 2 cos b
0
−
e
2
n
tan b cos l
−
e
2
n
tan b sin l
−
sin b cos l
m 2
−
sin b sin l
m 2
cos b
m 2
[m
2 −n
2 (1−e
2 )] sin l
m 3
[n
2 (1−e
2 )−m
2 ] cos l
m 3
· · ·
· · ·
0
0
0
0
0 −
e
2 n sin b cos b
m 2
e
2 n sin b cos b
a 1 m 2
[e
2 m sin
2 b+2(1−e
2 )n] sin b cos b
2m 2 (1−e 2 )
3
5 .
(21.45)
Example 21.1 numerically illustrates that for the computation of global {L, B} from local {l, b} and
from datum parameters {t x , t y , t z , α, β, γ, s, δa, δe
2
} we can neglect hB 1 . Accordingly, the synthesis of
{L, B} is performed by [L, B]
T = [l, b]
T + B 0 [t x , t y , t z , α, β, γ, s, δa, δe
2 ]
T .
Example 21.1 (Synthesis of {L, B} from {l, b} and from datum parameters {t x , t y , t z , α, β, γ, s, δA, δE
2
},
simulation of impact of local height h).
a 1 := 6 377 397.155 [m], e
2 := 0.006 674 372, A 1 := 6 378 137 [m], E
2 := 0.006 694 380 022 9.
l = 14
◦ , b = 40
◦ , h = 1000 m (assumed)
{t x , t y , t z , α, β, γ, s, δA, δE
2
} from first section.
L 0 = L(l, b)
L = L(l, b, h)
13
◦ 59
54
.269 6
13
◦ 59
54
.270 4
B 0 = B(l, b)
B = B(l, b, h)
40
◦ 00
00
.069 44
40
◦ 00
00
.069 42
End of Example.