Datum Problems - Map Projections: Cartographic Information Systems - page 523

Box 21.8 (Decomposition of the synthesis matrix B = B 0 + h B 1 +O 2 ( h 2 ).).

B=

=

e 2 ) n + h ]sin b cos l
( n + h )cos b

e 2 ) n + h ]sin b sin l
( n + h )cos b

[(1 − [(1 −

sin l
( n + h )cos b

cos l
( n + h )cos b

− 0 − −

··· (21.43)

[ n 2 (1 − e 2 )+ mh ]sin l
m ( m + h )

[ n 2 (1 − e 2 )+ mh ]cos l
m ( m + h )

sin b cos l
m + h

sin b sin l
m + h

cos b
m + h

− −

1 0 0 0

2 × 9 ,

··· ∈ R − [ e 2 m sin 2 b +2(1 − e 2 ) n ]sin b cos b

2( m + h )(1 −e 2 )

0 e 2 n sin b cos b
m + h

e 2 n sin b cos b
a 1 ( m + h )

− −

B 0 :=

=

sin l
n cos b

cos l
n cos b

e 2 )tan b cos l e 2 )tan b sin l

− 0 − (1 − − (1 −

··· (21.44)

[ n 2 (1 e 2 )] sin l

m 2

[ n 2 (1 e 2 )] cos l

m 2

− − −

sin b cos l
m

sin b sin l
m

cos b
m

− −

,

1 0 0 0

···

e 2 m sin 3 b cos b +2(1 − e 2 ) n sin b cos b
2 m (1

0 e 2 n sin b cos b

e 2 n sin b cos b
a 1 m

− −

m − e 2 )

B 1 :=

=

e n tan b cos l e n tan b sin l

sin l
n 2 cos b

cos l
n 2 cos b

− 0 − −

··· (21.45)

[ m 2 − n 2 (1 − e 2 )] sin l
m 3

[ n 2 (1 − e 2 ) − m 2 ]cos l
m 3

sin b cos l
m 2

sin b sin l
m 2

cos b
m 2

− −

.

0 0 0 0

···

[ e 2 m sin 2 b +2(1 − e 2 ) n ]sin b cos b
2 m 2 (1 −e 2 )

e 2 n sin b cos b
m 2

e 2 n sin b cos b
a 1 m 2

0 −

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 h B 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 . 006674372 , A 1 := 6 , 378 , 137 [m] ,
E 2 := 0 . 0066943800229 .
l =14 ◦ ,b =40 ◦ ,h = 1000m (assumed)

t x ,t y ,t z ,α,β,γ,s,δA,δE 2 { } from first section .

L 0 = L ( l,b )
13 ◦ 59 54 . 2696
B 0 = B ( l,b )
40 ◦ 00 00 . 06944

L = L ( l,b,h )
13 ◦ 59 54 . 2704
B = B ( l,b,h )
40 ◦ 00 00 . 06942

End of Example.