A Method for Calculating Exact Geodetic Latitude and Altitude ... - DTIC

NSWC TR 85-85

AD-A265 783

A METHOD FOR CALCULATING EXACT GEODETIC LATITUDE AND ALTITUDE

BY ISAAC SOFAIR

ED -CT fCz

STRATEGIC SYSTEMS DEPARTMENT

JUJN 2 1

3f993

APRIL 1985 REVISED MARCH 1993

Approved for public release; distribution is unlimited.

,*

NAVAL SURFACE WARFARE CENTER DAHLGREN DIVISION Dahigren. Virginia 2248-5000

93-13799

NSWC TR 85-85

A METHOD FOR CALCULATING EXACT GEODETIC LATITUDE AND ALTITUDE

BY ISAAC SOFAIR STRATEGIC SYSTEMS DEPARTMENT

APRIL 1985 REVISED MARCH 1993

Approved for public release; distribution is unlimited.

NAVAL SURFACE WARFARE CENTER DAHLGREN DIVISION Dahlgren, Virginia 22448-5000

NSWC TR 85-85

FOREWORD

The work described in this report was performed in the Fire Control Formulation Branch (K41), SLBM Research and Analysis Division, Strategic Systems Department, and was authorized under Strategic Systems Program Office Task Assignment 36401. This work was necessitated by the need to formulate an exact method of computing geodetic latitude and altitude. This report has been reviewed and approved by Davis Owen; Johnny Boyles; Robert Gates, Head, Fire Control Formulation Branch; and Sheila Young, Head, SLBM Research and Analysis Division. This document is a revised version of NSWC TR 85-85, dated April 1985. The distribution statement has been changed.

Approved by:

LR.I SC aIDT, Head Strategic and Space Systems Department

NTIS

M3

ulic

. I

A-Il

i'ii

J

M.~

I.....

1.

NS6C TR 85-R5

CONTENTS

page INTRODUCTION ......................

..............................

.

FORMULATION .......................

..............................

.

SOLUTIGX 6F BIQUADRATIC ......................

.......................

DERIVAT;ON OF GEODETIC ALTITUDE AND LATITUDE ..........

.............

3 7

CONCLUSION ............................

..............................

9

REFERENCES ............................

..............................

10

....................

APPENDIX--THE INVERSE PROBLEM ...............

.............................

DISTRIBUTION ..........................

A-I (1)

ILLUSTRATION

Figure 1

Pag~e ELLIPSOIDAL NORMAL ............

ii"/Iv

....................

8

I NTRODUCT ION

Fillowing is a simple and efficient model for calculating geodetic latitude and altitude of an arbitrary point in space,

the exact given the

coordinates of that point. The model assumes the earth to be an oblate spheroid; i.e., an ellipsoid of revolution whose semimajor axis (a) is the radius of the circle described by the equatorial plane, and whuse semiminor axis (b) is a line joining its center and one of its poles. The point in question can be either inside or outside the ellipsoid, but not too close to the center of the ellipsoid. The solution of the problem and its constraints will now be expounded.

FORMULATION

Choose an earth-fixed Cartesian coordinate system whose origin coincides with the center of the ellipsoid, The unit vectors i, j, k coincide with the (x, y, z) axes, respectively. The +z axis points in the direction of the North Pole. The +x axis is the line of intersection of the equatorial plane and the plane of zero longitude. The +y axis completes a right-handed coordinate system. An equation of the ellipsoid in

this frame is

z2

R2

1,

2

+

a2

where

R - Vx2

+ y2

Let P(xo, Yo, Zo) be the coordinates of the given point. It is desired to find the points P(x, y, z) at which the normals from P to the surface of the ellipsoid cut the ellipsoid. The given by -~

slope of a normal to a2 z

I

-•

2-

where

R2

--

+

the ellipsoid at any point on its surface z2

7-

(2)

.

dR Therefore the slope of a normal from P is z

Ro

-R

Z

a-z

b-R :

where Ra•0

t

is

NSWC TR

(Re

R)azz = (z 0

-

-

-•

z)bzR

or 2

Ra

z = R{a 2 z + b 2 (zo

-

z)j = Ri(a

2

-

bz)z + b½Zoj

Squaring both sides and expressing R in a 2 b2 R2z

2

- (b2 -

z2 )j(at

Writing the above following biquadratic in b 2 ) z" + 2b'(a

-

2 b(a

2

2

-

in

b2 )z 0 z3

2

+ blz01

descending

powers of

+ b2 {a2 Rý + b 2 z2 -

(a

2

z,

-

h

we obtain the

2-

b 6 z2 = 0.

b 2 )z 0 z -

-

b 2 )z

equation z:

2

(az

-

terms of z,

Now z is either positive or regative and therefore is not an appropriate variable for the solution. However, this difficulty is easily circumvented by introducing a new dimensionless variable k * Z Using the constraint that z -.

zo

always has the same sign as z0 , we see that k is always positive. Putting z z 0 k in the biquadratic and writing the resulting equation such that the coefficient of k' is unity, we finally obtain k4 + 2pk 3 + qk 2 + 2rk + s

0

where

b2 p

7a

2

,b2 + b 2 z2

b 2 {a 2 R,

(a

q

2

-

-

(a2

b2)2z2

-

2)1

(4)

b"u

r

=

-

(a2

b 2 )z25

-

6

2

(a

b 2 - b2 ) 7

Now we will employ two of thrpo powprful theorems in the Thpory of Fqu"itions to expose the nature of the roots of (*). The third theorem Vill hP needed later on. The proofs of these theorems are given in Reference I.

NSWC TR

I. An equation f(x) changes of sign in f(x) , changes of sign in f(-x).

b-?

roots Than there ;r,roots than there arp

have more positive have more negative

= 0 cannot and cannot

II. Every equation which is of an even degree and has its negative has at least two real roots, one positive and one negative.

is

III. Every equation of an odd degree has at last term. opposite to that of its

By observing conclusions: (i) two real

(*)

and (3),

(5),

(6),

we deduce

Since the last term of (*), namely s, roots of opposite sign (using IT).

(ii) Since there whether q is positive (using I).

root. seek.

(4),

least one real

is

is

tern

root whose sign

at onre

negative,

last

the following

there are at

only one change of sign in (*), irrespective or negative, there is at most one positive

least

of root

From (i) and (ii) we see immediately that (*' has exactly one positive Since k is constrained to be positive, this positive root is the one we

SOLUTION

OF BIQUADRATIC

The solution of (*) is effected method which will now be enunciated. In

by a standard method known as Ferrari's

the equation

k1 + 2pk

3

+ qk 2 + 2rk + s

- 0

add to each side (ck + d)z, the quantities c and d being determined make the left-hand side a perfect square; then k4

+ 2pk

3

+ (q + c 2 )k 2

+ 2(r + cd)k + s + d2

a (ck

Suppose that the left side of the equation then by comparing the coefficients, we have p 2 + 2t = q + C2, by eliminating (pt

-

or 2t

r)} 3

-

= (2t

qt

2

pt - r

+ cd,

+ 2(pr

-

-

2

q}{t

s)t

-

-

+ d)2 . equal

t' = s + d 2 .

c and d from these equations, + p2

is

S

so as to

to (k 2 + pk + t)2,

(7) we obtain

,

p 2 s + qs 3

r2

= 0.

(8)

NSWC TR A;-85

From (3),

(4),

(5),

and (6),

we see that 2 a b8 R2

pr

0, -p

s =

-

2

s + qs -

r

Substituting into (8),

2

= -

-

we obtain the following cubic in

t:

az b R2 2t'

-

qt

2

(a

-

2

bzPzz

-

we see that

to (8'),

III

Theorem

Applying

0.8')

=

root, irrespective of whether q is exactly one positive root. The solution of a cubic Cardan's Solution. First eliminate the t

t

2

is

has at

(8')

least one positive

has at most one positive

positive or negative.

accomplished

term in

(8')

we see that it

Now applying Theorem I to (8'),

root.

Hence

:8')

has

by a standard method known as

by making the substitution

_ t, +_q

6

i.e., a? b8 R 2(t" + 2)3

-

+ 2)2

q(t'

-b

6(a2

2

-0

)4

or

!32-

t-3

12

a 2 bRS

108

2(a

2

b 2•z•

-

0.

Let

f=

a

S3 h = CL

12

108

a2b8R 2(a

2

-

b 2 )'z'

so that the above equation can be written in t'3

+ ft

+h

To solve (9), t

3 =

u + v)

the form

0. lot

(9) r' = u + v, then

= u+

+

3

uv (u

+

4

uU)3

+ v,

+

uvt"

~~

NSW(C Tk and (9) becomes U& + V3 + (3uv At their satisfy

+ h = 0.

+ f)t'

present, u and v are any two quantities subject to the condition thar sum is equal to one of the roots of (9); if we further suppose that th-v We thus 0 0, they are completely determinate. the equation 3uv + f

obtain U3 + V

h u 3 v3 h,

=

-

27 u3 , v 3 are the roots of the quadratic

Hence, •z

+ ha

C-

h

f3 -

=

Thus,

h2 +"-

Putting u 3

h42

h"

3 f 2+7

h 2 + f- 3

2-h

v3

,we

obtain t

from

u + v.

the relation t' Hence,

f3

h2

t

'

•

"

hf

7

f3

ic valid provided that h-- + f

The soýltinr (10)

"(10)

Ž>0.

It

will now be

shown that this constraint is valid for all points of interest. a~b8 R• -+ h2

f,

I

727

Z

2(a2

108 3

a 2 b 8 R2q

"432(a 2

-

a2b"'Ra

R 4

432(a 2

-

16(a

+

+ blz 32(a

h)1O

2

-

46656

Z.

a'b

b 2 )4z 2

=

2 y-

6

)

_

____q

___a_2 '-

2

1 6 R•

b 2 ) 8 zo

-

(a2

-

+ 16(a 2

_ b 2 ) 1 0 z•O

R

+

h2 z

27a 2 h 2 (a2 -

• m5

.

a4b'

b2

-

_ bZiazg,

b))aV a

(a + -

W ) 2 Rpz20

6

I

+

using (4

cc,

the required

Thus,

{a2 R4

+

b-'h 2

(a2

7, b"~

27ah-'( a2

+

-'

h-'

>~~~(

s it is seen that rho above ronst ra ,ne By inspect ion and a Iso by trial, ir whpre points only the fact, of matter a As points of interest. valid at all rho i.e., closp to the orijn, are relatively does not hold are those that inrprost. practirAl no of and these points are center of the ellipsoid; Now

h

a 2 ba R2

-

-

108

2

2Da

b6 ýa2

h 2,

-

z;

+ b 2zZ2

N2

108(a

2

a2

a2

2 )'RV

b 2 )b

-

2(aab

b'

-

)'

z•

b 62 1o8(a

---

2

-

If the constraint Hence, h is positive.

(9)

has at

[1a 2

b)Z

-a 2

+ b 2z•

-

I

bh

holds, then the expression (11) Applying Theorem III negative.

4a 2

+

-

.

in the brackets abovP is we conclude that to (9),

Now applying Theorem I to (9)

least one positive root.

h )R-z

and using

both f and h are negative, we conclude that (9) has at most onp that the fact It will now he shown has exactly one positive root. Hence, (9) positive root. that this positive root is given by (10). h2 Let

g

fV

+ T7

-

,

then

2

Now

g

(f) f3-
v

- -

> 0O.

12

12

0

< 0, which 0, % 0

since h Thus,

hc

t

=

mlr

ht

impll .a h

thit

_;

u + v > 0.

>

V

h

-h

V-7 h

+ T

Hence,

and

V/g

+ h

h-

\;W( Prore-d ing v ith nour derva t inn, t

"+

Tk-'-," ý;o hArp

> 0

q b

Solving foi c and d from the svstpm of equations (7), c - ý p 2 - q + 2td Now (kV or k2

2

t

-s

( ck + d

+ pk + t

(ck

+ pk + t

,

+ d)

from which we obtain the two quadratics

k+

c)k + t - d

+ (p -

in

k:

0

+ c)k + t + d = (t)

k2 + (p

-P

we obtain the following four roots of

and (13),

From (12)

) +

--

(p

-

2

C)

2

4(t

-

d)

-

4(t

-

d)

+ d)

-(p

c)

-

4(P

-c)

-(p +

C)

+

ý(p

+ c)-'

-

4(t

2

-

4(t + d)

k22

we havr

*(:

k3 =2 -(P

+ c)

- ,4(p + c)

DERIVATION OF GEODETIC ALTITUDE AND LATITUDE

Since It has already been shown that (*) has exactly one positive root. t2d 2 = s < 0, (t + d)(t - d) < 0, which implies that t + d and t - d are of Now t - d ir the last term opposite sign; but d > 0; thus t + d > 0 > t - d. Hence, by applying Theorem II of (12) and has just been shown to bp negative. must contain the desired positivo root. we see immediately that (12) to (12), it is clear that k, is the required two roots above, Inspecting the first positive root.

The rest of the solution follows trivial.Iv. 7

=

•"

Let k = k!.

th-n

NStýC TV~bS~ R -a

X

0 X0

1

-

R

,

and

-

'0 RU

Stan1

x (Here

k is

the longitudp at (x,

y)).

The above solotion is not applicable to the case 7. 0 since two of tho baeic parameters of the solution, q and s, involve a division by z 0 . Also, thrý

sol ition is not applicable to the case R. - 0. Hence these two cases havp to he created separately; nevertheless, this poses no difficulty since the risulrs are already known then. The geodetic altitude is

given by

R 2z01.2

H = sign

= sign

12+ aa22

a2

b2 2

b2

-

-

10,-

xt _-Xo x 2+ý"_Y2 2 +( 0

1

(

0--

R 2 + (z

Z)_Z1 z

5)

0-Z)

In order to find the geodetic latitude and the unit normal vector at (x, y, z), it is desirable to introduce a geometric term, Ne, which is never zero. Ne is defined to be the distance along the ellipsoidal normal from the surface of the ellipsoid to the z-axis (Reference 2). (See Figure 1.)

R

FIGURE 1.

ELLIPSOIDAL NORMAL

NSWC TR 85-85

From Figure 1, we have R

Ne

Also, a

tan

2

z

b-R 2

from (2).

Hence, 4

cos c

sin Thus,

1

R sec 0 = R

Ne

4 tan 0

tan

+

aý2

4

2

-

1

R

+ h' R2

V.--,2

-R 2 b7R

+

b2b2

=

2

a2 z b 2 Ne

the components of the unit normal vector at (x,

y,

z)

are

2

HHx Ne

(a)~ z

yH

Ne

'

Ne

The geodetic latitude 0, is

S= sin- 1 H 3 This completes our solution.

CONCLUSION

In contrast to other methods currently in use, this method theoretically computes the exact values of the geodetic latitude and altitude, assuming the The method was programmed and earth to be a perfect ellipsoid of revolution. run

It

for many given values of x 0 , yo , z.

gave results which are consistent

with those of other methods, including Brookshire's approximation (Reference 3) For Roundoff errors are negligible. and an iterative technique (Reference 4). all practical purposes, the results are excellent and the method is definitely recommended as an alternative approach to problems involving geodetic latitude.

9

NSWC TR 85-85

REFERENCES

1.

H. S. Hall and S. R. Co., London, 1927.

2.

Davis L. Owen, Geometry of the Ellipsoid, Dahlgren, Virginia, December 1984.

3.

J. W. Brookshire, A Direct Approximation Method for Calculating Geodetic Latitude, Point Mugu, California, 1981.

4.

Weikko Heiskanen and Helmut Moritz, Geodesy, Graz, Austria, 1979.

Knight,

Higher Algebra,

Fourth Edition,

Naval Surface Weapons Center,

Physical Geodesy,

1(

Macmillan and

Institute of Physical

NSWC TR 85-8

APPENDIX THE INVERSE PROBLEM

NSWC TR 85-,'

APPENDIX THE INVERSE PROBLEM

An arbitrary H 2 , H 3 ), (x, v,

point in space is zo ) z), and (x 0 , yo,

relatively easy to solve.

This problem is R - Ne cos

4, z =

a2

Using the relation a

z2 +

1, we obtain

12

+ h-

sin2o C,

1;

or a

a

Ne

sin2

cos 2 o + 1

Also,

from page 8,

X - tan-'

x x

-_(1

-

b)

we have sin- 1 I

cos-I R

x

R

Therefore, x - R cos X a Ne cos 4 cos X, y = R sin X -

Ne cos 0 sin A.

Hence,

[x

I Izi

rNe

cos 0cosI

Ne cos i$sin X a , Ne sin

A-i

k, 0.

and

From page 9,

b22

Nea2 (cos2,

by

Ne sin 0.

--

R2

a2

defined

sin 2 o

H.

Solve

we have

for (HI,

H=

Cos\

Cos

H

ssin

cos

Ne [H3 (a 2z bx

x°

siL

Te ,

that From Figure 1, it follows immediately

rxO [o

[zo

1y

[(Ne + H) cos 0 cos H, 1 H,= (Ne + H) cos 0sink H

([2-

Ne + H)

sin

A-2

NSWC TR 85-85

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3. REPORT TYPE AND DATES COVERED

2. REPORT DATE

March 1993 4. 1TInE AND SUBTITLE

S. FUNDING NUMBERS

A Method for Calculating Exact Geodetic Latitude and Altitude

6. AUTHOR(S)

Isaac Sofair 8. PERFORMING ORGANIZATION

7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES)

REPORT NUMBER

Naval Surface Warfare Center (K41) Dahlgren Division Dahlgren, VA 22448-5000

NSWC TR 85-85

9. SPONSORING/MONITORING AGENCY NAME(S) AND

10. SPONSORING/MONITORING AGENCY REPORT NUMBER

11. SUPPLEMENTARY NOTES

12a. DISTRIBUTION/AVAILABILITY

12b. DISTRIBUTION CODE

Approved for public release; distribution is unlimited.

13. ABSTRACT (Maximum 200 words)

This report derives a new and improved technique for computing the geodetic latitude and altitude of an arbitrary point in space assuming the Earth is an ellipsoid of revolution. This method theoretically computes the exact values of the geodetic latitude and altitude. All the programs currently in use utilize approximation techniques for computing the quantities mentioned above. The method is simple and efficient and is strongly recommended for implementation.

14. SUBJECTTERMS

15. NUMBER OF PAGES

ellipsoid of revolution, exact geodetic latitude, exact geodetic altitude, roots of

biquadratic equation 17. SECURITY CLASSIFICATION OF REPORT

UNCLASSIFIED NSN 7540-01-280-5500

21 16. PRICE CODE

18. SECURITY CLASSIFICATION OF THIS PAGE

UNCLASSIFIED

19. SECURITY CLASSIFICATION OF ABSTRACT

20. LIMITATION OF ABSTRACT

UNCLASSIFIED Stanaard form. 298 (Rev 2 89)