Survey Review XXIII, 181, July 1976

TRANSFORMATION FROM SPATIAL TO GEOGRAPHICAL

COORDINATES

B. R. Bowring

Directorate of Overseas Surveys

ABSTRACT

Formulae relating cartesian to spheroidal coordinate systems are examined. A simple equation
eliminating the usual need for iteration is presented. Tests show that this formula is virtually
rigorous for terrestrial stations.

NOTATION

a semitnajor axis of spheroid

b sen1itninor axis of spheroid
e2 I-b2/a2
e a2/b2-1
¢ geographical latitude
II paran1etric latitude
A longitude
v radius of curvature in the prin1e vertical for latitude ¢
H spheroidal height
X YZ cartesian coordinates

THE PROBLEM

The proble111 is to deternline ¢, A, H fro111given (X, 1': Z). The longitude }.

is given at once by the fonnula:
tan A = Y IX. (1)
Finding ¢ and H presents some difficulty. By syn11netry it should be noted that
the problem for ¢ and H can now be reduced to the case corresponding to the
meridional plane through the given point P. Let the X-axis lie in this plane, so
that the cartesian coordinates becon1e (X, Z). In the general case X should be
replaced by J (X2 + y 2).

ITERATIVE SOLUTION

In Fig. 1, OG =e 2
v. sin ¢ so that in triangle PGN

~ Z + e2 v . sin ¢
tan,+, = X . (2)

This can be used to obtain ¢ by. iteration using a starting value of ¢ such as fron1

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 TRANSFORMATION FROM SPATIAL TO GEOGRAPHICAL COORDINATES

z
x
N

o
x
c

Fig. 1

tan 4>= (1 + 8)(Z/ X). This value of 4> is for the point Po where PO cuts the
spheroid.

Putting H = Q.e2 v,

X = (1 + e2 Q) v cos 4>
so
tan 4>= (Z/X)(1 +e2 Q)/[1 +e2(Q-l)].

Now e2 v is approximately 43 km and choice starting values would be from:

tan 4>= (Z/X)(l +8) surface (Q = 0) (3.1)

tan4>=(Z/X)(I+e2) at43km (Q=I) See [1], page 5 (3.2)

2 2
tan4> = (Z/X)(1 +e Q)/[1 +e (Q-l)] at 43Q km when Q is known. (3.3)

In the absence of any knowledge of the value of H, the equation (3.2) would
appear to be the most sensible for earth bound stations.

THE PROPOSED NEW FORMULA

The centre of curvature C of the spheroid corresponding to P' (the foot of the
normal P P') is the point

(e2 a.cos3 u, -e.b.sin3 u),

where u is the parametric latitude of the point P'. Therefore (see Fig. 1)

A" Z + e . b . sin 3 u
tan'f'=----- (4)
X -e2 a.cos3 u
This is clearly an iterative solution; but it has been found that this formula is
extremely accurate on test using the single first approximation for u for the right-
hand side from tan u = (Z/X) (a/b).

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 B. R. BOWRING

EXAMPLE 1
Spheroid: a = 6378249·145 m e2 = 0·0068035111
b = 6356514·870 m e = 0·0068501160
Given point: X = 4114496·258 m Z = 4870157·031 m
Solution: tan u = (a/b)(Z/X) = 1·1877053292
cos u = 1/v"(1 +tan2 u) = 0·6440705847
sin u = v"(I-cos 2
u) = 0·7649660659
Then by (4) tan if1 = 1·1917535925
if1 = 50°00'00·0000"
The formula will now be proved analytically.

PROOF OF FORMULA AND DERIVATION OF FURTHER TERM

By (2), noting that (l-e2)v.sin¢ = b.sinu, and putting tan 8 = (Z/b)/(X/a),

tan u-tan8- 2
(e a/X) sin u+e 2
tan 8 = o.
Differentiating with respect to (tan u),
d(sin u)/d(tan u) = cos3 u.
Hence
(e2 a/X) sin8-e2 tanO
tan u = tan 0 + -------- 2
1- (e a/X) cos3 0

tan 0 - (e2 a/X) sin O.cos2 0 + (e2 a/X) sin 0 - e2 tan 0

1- (e2 a/X) cos3 0 -

(a/b)(Z/X) + (e2 a/X) sin3 0- (l-b2/a2)(a/b)(Z/X)

1- (e2 a/X) cos3 0

b Z + e . b . sin 3 0
- -;; X _e2 a.cos3 O·

Formula (4) then follows because tan ¢ = (a/b) tan u. Using now the next
term of Taylor's Series,
tan u = tan 0
(e2 a/X) sin O-e2 tan 0
+ -------------------------
1
1- (e2 a/X) cos3 0 + 2! [3(e2 a/X) cos4 0 sin O][(e2 a/ X)sin 0 - e2 tan OJ

So
b Z+e.b.sin30
tanu = -. ----- -At
a X -e2 a.cos3 0
where

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 TRANSFORMATION FROM SPATIAL TO GEOGRAPHICAL COORDINATES

TABLE 1

H in metres Maximum D.eP"

- 5 OOO}Earth 0·0000
a bound 0·0000
+ 10 000 region 0·0000 ~ Maximum 0·000 000 030"
400 000 0·0000 in earth bound region
500 000 0·0001
1 000 000 0·0002
2 500 000 0·0007
5 000 000 0·0013
10 000 000 0·0018 ~ Maximum o· 0018/1 at H = 2a
20 000 doo 0·0017
30 000 000 0·0014
100 000 000 0·0006
500 000 000 0·0003
1 000 000 000 0·0001
2 000 000 000 0·0000
00 0·0000

Now

Then
!J.t = +i[(R -1)2/R3] e6 sin30 cos O.
With enough accuracy for !J.t,
R = OP/a = (H+a)/a,
and 0 may be replaced by ¢. So
!J.t = +1-e6 a[H2/(a+H)3] sin3 ¢ cos ¢. (5)
It follows that the corresponding error in latitude will be
!J.¢ = +1-e6[aH2/(a+H)3] sin3 ¢ cos3 ¢
!J.¢" = +1-e6 [206265 aH2/(a+H)3] sin3 ¢ cos3 ¢
} (6)

EXAMPLE 2

a = 6·38 106111 e2 = 0.0068 H = 10000m ¢ = 500

!J.t = 0·000 000 000 000 33 !J.¢" = 0·000 000 028"

Now put F = H2/(a+H)3.

Then dF/dH = [2H(a+H)3-3H2(a+H)2]/(a+H)6.
For H -:f. 0, dF/dH = 0 when H = 2a.
Also sin ¢ cos3 ¢
3
= tr sin3 2¢ ~ -!.
So by (6) A¢" ~ 0.0018".

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 n. R. nOWRING

Let x = (H in Inetres)/I06• Then the lnaxinulm A4>" for height H is given by

lnaximulll A4>" = 0'078x 1(6· 38 + X)3.
2
(7)
For earth bound points the maxilnum is not lTIOre than 0·000 000 030".
Table 1 shows the values of maximum A4>" for different heights as given by
formula (7). Should greater accuracy be required in the region of distant space,
formula (4) should be used with a single iteration noting that tan 11 = (bla) tan 4>.

HEIGHT

Once the latitude ¢ has been determined the height is given by the closed
formulae:
1-1 = X sec4>-v = Z cosec4>-(1-e2) v. (8)

Reference
1. R. A. Hirvonen and H. Moritz (1963): Practical Computations of Gravity at High iltitudes
Report No 27. Inst. Geod. Phot. Cart. Ohio State Univ.

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