458 21 Datum problems
Continuation of Box.
Curvilinear coordinate conformal transformation close to the identity
(datum transformation).
1st variant:
l =
= arctan
“
Y + t y + Zα − Xγ + Y s
”
/
“
X + t x − Zβ + Y γ + Xs
”
,
b = b(X, Y, Z, t x , t y , t z , α, β, γ, s) .
(21.18)
2nd variant:
tan l =
=
“
[A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B sin L + t y
+ [A 1 (1 − E
2 )/
p
1 − E 2 sin
2 B + H(L, B)] sin Bα
− [A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B sin Lγ
+ [A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B sin Ls
”
/
“
[A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B cos L + t x
− [A 1 (1 − E
2 )/
p
1 − E 2 sin
2 B + H(L, B)] sin Bβ
+ [A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B sin Lγ
+ [A 1 /
p
1 − E 2 sin
2 B + H(L, B)] cos B cos Ls
”
,
l = l(L, B, H(L, B), t x , t y , t z , α, β, γ, s, A 1 , E
2 ) ,
b = b(L, B, H(L, B), t x , t y , t z , α, β, γ, s, A 1 , E
2 ) .
(21.19)
Taylor expansion:
l =
= L + l t x t x + l t y t y + l t z t z + l α α + l β β + l γ γ + l s s + l A δA + l E δE
2 ,
b =
= B + b t x t x + b t y t y + b t z t z + b α α + b β β + b γ γ + b s s + b A δA + b E δE
2 ,
(21.20)
subject to
l t x :=
∂l
∂t x
(t x = t y = t z = 0, α = β = γ = 0, s = 0, A 1 = a 1 , E
2 = e
2 )
etc.
(21.21)
and
δA := A 1 − a 1 , δE
2 := E
2 − e
2 , δE = (2E)
−1 δE
2 .
(21.22)