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where e 2 =( a 2
b 2 ) /a 2 . Dividing this equation by the above expression for
p , we find
= 1 − e 2
tan ϕ,
Z
p
N
N + h
(5-31)
so that
1 − e 2
1
tan ϕ = Z
p
N
N + h
.
(5-32)
Given X, Y, Z , and hence p , Eqs. (5-29) and (5-32) may be solved iteratively
for h and ϕ . As a first approximation, we set h = 0 in (5-32), obtaining
tan ϕ (1) = Z
e 2 ) 1 .
p (1
(5-33)
Using ϕ (1) , we compute an approximate value N (1) by means of (5-21). Then
(5-29) gives h (1) .Now,asasecondapproximation,weset h = h (1) in (5-32),
obtaining
1 − e 2
1
N (1)
N (1) + h (1)
tan ϕ (2) = Z
p
.
(5-34)
Using ϕ (2) , improved values for N and h are found, etc. This procedure is
repeated until ϕ and h remain practically constant.
The result for λ is immediately obtained from the first two equations of
(5-27):
λ =arctan Y
X .
(5-35)
Many other computation methods have been devised. One example for
the transformation of X, Y, Z into ϕ, λ, h without iteration but with an
inherent approximation is
ϕ =arctan Z + e 2 b sin 3 θ
p
,
e 2 a cos 3 θ
λ =arctan Y
(5-36)
X ,
p
cos ϕ
h =
N,
where
θ =arctan Za
pb
(5-37)
is an auxiliary quantity and
e 2 =( a 2
b 2 ) /a 2 ,
e 2 =( a 2
b 2 ) /b 2
(5-38)
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