SUG456-4N.pdf

LECTURE 4 SUG/GLS456 GEODESY

PPSUG, FSPU, UiTM, SHAH ALAM

ASSOC. PROF. Sr. DR. AZMAN BIN MOHD. SULDI (MRICS, MRISM)

ASSOC. PROF. PROF. Sr. Sr. DR. AZMAN BIN MOHD. SULDI PPSUG, FSPU, UiTM, SHAH ALAM

SUG456 GEODESY : ELLIPSOID COMPUTATIONS

CALCULATION OF GEODETIC COORDINATES (SOLUTION OF THE ELLIPSOIDAL POLAR TRIANGLE) The geodetic coordinates of points on the ellipsoid are usually specified as Latitude and Longitude. If we assume that we are given the coordinates of a starting point, the distance and azimuth to the 2 nd point, we desire to compute the coordinates of the 2 nd point as well as the azimuth from the 2nd point to the 1st. Such problem is defined as the direct geodetic problem or simply the direct problem. (i.e. given φ1, λ1, s12, α12 : calculate φ2, λ2, α21) The inverse geodetic problem is defined as the case where the coordinates of the end point of the line are given and we desire to find find the azimuth from point one to point two, the azimuth from point two to point one, and the distance between the two points. (i.e. given φ1, λ1, φ2, λ2, : calculate, α12 , α21, s12) The solution of either of these problems is basically the solution of the ellipsoidal polar triangle shown in Fig 7.1



CALCULATION OF GEODETIC COORDINATES (SOLUTION OF THE ELLIPSOIDAL POLAR TRIANGLE) Pole

Δλ

We can expressed express ed the defined problems in the following functional forms Direct problem : φ 2 = f 1 (φ 1, λ 1, α 12 , s) λ 2 = f 2 (φ 1 , λ 1,α 12 , s)

(7.1)

α 21 = f 3 (φ 1, λ 1,α 12 , s )

P1

(φ1, λ1) α12

P2

s12

α21

(φ2, λ2)

Inverse problem : s = f 4 (φ 1 , λ 1 ,φ 2 , λ 2 )

Fig 7.1 The Polar Polar Ellipsoidal Triangle

A minimum of 3 independent formula are required and often given in the following form;

a12 = f 5 (φ 1, λ 1 ,φ 2 , λ 2 ) α 21 = f 6 (φ 1, λ 1,φ 2 , λ 2 )

Δφ = φ 2 − φ 1 = ....... Δλ = λ 2 − λ 1 = ....... Δα = α 21 − α 12 −180o = .......

(7.2)



CALCULATION OF GEODETIC COORDINATES (SOLUTION OF THE ELLIPSOIDAL POLAR TRIANGLE) Pole

Δλ

P1 (φ1, λ1)

P2

α12 s12

α21

(φ2, λ2)

Fig 7.1 The Polar Polar Ellipsoidal Triangle

There are many solutions for these problems. Such solutions are generally classified by the distance for which they are valid and by the type (i.e. normal section or geodesic) of the line being considered. Solution for long lines require more extensive formulas. Some authors group these formula under one generic name – Long Long Line Formulae. Formulae. Others differentiate between Short Line, Line, Medium Line and Long and Long Line Formulae depending on their applicability. applicability.

Also, due to many formulae available, they are usually categorized into two categories depending on their mode of derivations. i.e. Normal Section Formulae – Formulae – those formulae where their derivation is dependent on an osculating sphere. Geodesic Formulae – Formulae – Those which utilizes the differential differential equations of the ellipsoidal geodesic.

ASSOC. PROF. Sr. DR. AZMAN BIN MOHD. SULDI PPSUG, FSPU, UiTM, SHAH ALAM


CALCULATION OF GEODETIC COORDINATES (Normal Section Formulae) These are usually derived by taking a sphere which touches the ellipsoid at some convenient point, such as the stand point or mid-point of the line or at the equator. Among the best known of these are the Clarke-Robbins formulae and Puissant’s formula. The simplest, conceptually, are the Gauss Mid-Latitude formulae (GML). In deriving these formulae, Gauss mapped the ellipsoid conformally on to a sphere of radius R imposing the conditions that (i) the scale at a particular latitude φ o should be true. (ii) the variation in scale away from this latitude should be as little as possible. . o , then for points within 1o (100km) of φ o , the ellipsoidal He found that if R = M o N angles and distances were mapped on to the sphere without distortion.

Gauss therefore conformally mapped the ellipsoidal φ1, λ1, α12 and s on to the sphere, solved by spherical trigonometry for φ2, λ2 and α21, and then transferred these back to the ellipsoidal surface.



CALCULATION OF GEODETIC COORDINATES (Normal Section Formulae) - GML As a result, he obtained the following formula for terms; s. cosα m ⎛ Δλ 2 Δλ 2 . sin2 φ m ⎞ ⎜1 + ⎟⎟ Δφ " = + 24 M m . sin1" ⎜⎝ 12 ⎠

Δλ " =

s.sinα m . secφ m N m . sin1"

Δα " = Δλ ".sinφ m Where;

Δφ = φ 2 − φ 1

Δφ, Δλ, Δα, neglecting the 5th order

⎛ Δλ 2 .sin2 φ m Δφ 2 ⎞ ⎜⎜1 + ⎟⎟ − 24 24 ⎠ ⎝

(4.11)

⎛ Δλ 2 .sin2 φ m Δλ 2 . cosφ m Δφ 2 ⎞ ⎜⎜1 + ⎟⎟ + − 24 12 12 ⎠ ⎝ in radian measure, or Δ φ " in seconds of arc

Δ = λ 2 − λ 1

in radian measure, or Δλ " in seconds of arc

Δα = α 21 − α 12 ± 180o

in radian measure, or

Δα "

in seconds of arc

It is obvious, the solution of the direct problem requires iteration. But not so for the inverse problem.



CALCULATION OF GEODETIC COORDINATES (Normal Section Formulae) - GML Pole PC

Δλ

A simple derivation of the Gauss formulae may be achieved if we consider a sphere which touches the ellipsoid at the point of the line between 1 and 2 where the latitude 1

= (φ 1 + φ 2 ) = the mid- latitude 2

α12+Δα α12

s

1A

(φ1, λ1) Fig 4.3

Using the standard formula of spherical trigonometry, one can find the relationship between the elements of the spherical triangle P12. (Figure 4.3)

2B

(φ2, λ2)

It is obviously advantageous to obtain values Δφ = φ2−φ1, Δλ=λ2−λ1 etc. In order to minimize the number of significant figures needed for a certain required accuracy. Hence, we shall use the half-angle formulae of Delambre and Napier.




Δλ

The convergence Δα , is deduced by straight substitution in one of Napier’s anologies, obtaining tan

1 1 ⎞ 1 Δα = tan Δλ .sin⎛ + Δ Δφ φ φ . sec ⎜1 ⎟ 2 2 ⎝ 2 ⎠ 2

1

(4.12)

Δα is of course the convergence on the sphere, α12+Δα α12

s

2B

(φ2, λ2)

but a theorem due to Dalby has shown that convergence on the conformal sphere is equal to convergence on the ellipsoid with negligible inaccuracy for lines as long as 800km. Also, since Δα, Δφ, and Δλ are small angles, it is sufficient to only use the first term of the series

1A

(φ1, λ1) Fig 4.3

Δα " = Δλ ".sinφ m Δα " = Δλ ".sinφ m

⎛ Δλ 2 .sin2 φ m Δλ 2 . cosφ m Δφ 2 ⎞ ⎜⎜1 + ⎟⎟ + − 24 12 12 ⎠ ⎝ (4.13)




Δλ

Δφ , is obtained by simple substitution into one of Delambre’s anologies, obtaining sin

α12+Δα α12

s

1A

(φ1, λ1) Fig 4.3

2B

(φ2, λ2)

1 2

Δφ = sin

s 2 R

⎛ ⎝

. cos⎜α 12 +

1 Δα ⎞⎟.sec Δλ 2 ⎠ 2

1

s Δφ Δλ Again, if we assume 2 , 2 and 2 R are small angles, using the first term, we have s 1 ⎞ Δφ = . cos⎛ ⎜α 12 + Δα ⎟ R 2 ⎠ ⎝

The problem of transferring this spherical Δφ back to the ellipsoid is reduced if we make R equal to M m, the radius of the curvature of the ellipsoid of the meridian at the mid-latitude. Then, converting Δφ to seconds of arc by multiplying it by 206265 or (1/sin 1”) we obtain, s. cosα m s 1 ⎞ ⎛ . cos ⎜α 12 + Δα ⎟ or = (4.14) Δφ " = M .sin1" 2 M . sin 1 " ⎝ ⎠



CALCULATION OF GEODETIC COORDINATES (Normal Section Formulae)- GML Pole PC

Δλ

Δλ , is obtained by straight substitution into Delambre’s formula, giving sin

1 ⎞ ⎛ 1 ⎞ s Δλ = sin⎛ ⎜α 12 + Δα ⎟.sec⎜φ 1 + Δφ ⎟.sin 2 2 ⎠ ⎝ ⎝ 2 ⎠ 2 R

1

And, if we use the first term only in the series of expansion of small angles Δλ and s , we 2 2R obtain

α12+Δα α12

s

1A

(φ1, λ1) Fig 4.3

2B

(φ2, λ2)

1 ⎞ Δα ⎞⎟.sec ⎛ ⎜φ 1 + Δφ ⎟ R 2 ⎠ ⎝ 2 ⎠ We can, by the same argument as for Δφ, use N m for

Δλ =

s

⎛ ⎝

. sin ⎜α 12 +

1

R, and converting to seconds of arc, obtain

Δλ " =

s N m . sin1"

⎛ ⎝

.sin⎜α 12 +

Δα ⎞⎟.secφ m 2 ⎠

1

Which is the first term in Gauss formula.

(4.15)



USE OF THE GAUSS MID-LATITUDE FORMULAE

Their use can be illustrated by reference to the short formulae, which are collected here, s (i) Δφ " = . cosαm (4.14) M m .sin1" s (ii) Δλ " = .sinα m .secφ m (4.15) N m .sin1" (iii)

Δα " = Δλ ".sinφ m

A. The Direct problem.

Given

(4.13)

φ1, λ1, s12, α12 : find φ2, λ2, α21

First iteration,

Δφ " =

s

. cosα12 M 1. sin1" s Δλ " = . sinα 12.secφ 1 N 1. sin1"

Δα " = Δλ ".sin φ 1



USE OF THE GAUSS MID-LATITUDE FORMULAE

Second iteration : Use correct formulae with values determined from 1 st iteration, and continue iterating until the difference between successive values of Δφ, Δλ and Δα is within the accuracy requirement. Three iterations usually suffice when using (4.13), (4.14) and (4.15). B. The Inverse problem.

Given

Divide (4.15) by (4.14), we have

Where everything is known except Then, obtain

Δα”

φ1, λ1, φ2, λ2:

find s12, α12 and

α21

Δλ M m = . tanα m . secφ m Δφ N m αm

from (4.13) and s from (4.14).

Hence, NO ITERATION is necessary. An accuracy of 1ppm is achievable with formulae (4.13), (4.14) and (4.15) for distances up to 30km, with (4.11) up to 100km.



7.2 Clarke’s formula for Geodetic Line (Normal Section Formulae) Pole P

Δλ

Let s = distance between point A and B Az = the azimuth from point A to B In plane Cartesian coordinate system

Δ x = s sin A z

;

Δ y = s cos A z

If we think of s cos Az as distance along the meridian (i.e. M Δφ) and s sin Az as distance along parallel of latitude (i.e. pΔλ= N cos φ Δλ ) we can write : B s

(φ2, λ2)

A

(φ1, λ1) Fig 7.2 The geometry of the development of the Clarke’s Formula

s cos A z M s sin A z N cosφ

= Δφ rad (change

in latitude)

= Δλ rad (change

in longitude)




ω

OR, in seconds of arc,

Δφ =

Δλ = K

s cos A z M sin1" s sin A z N cosφ sin1"

Where, N = the radius of curvature in the prime vertical M = the radius of curvature in the meridian Q

For simplicity, first consider the problem on the sphere Az

B s

(φ2, λ2)

Draw a line from B to line AP at right angle and draw arc BK such that its length = ω (i.e. difference in longitude)

A

(φ1, λ1)

ε = spherical excess in triangle ABQ (AreaΔ) =

Fig 7.2 The geometry of the development of the Clarke’s Formula

Area Δ R2




ω

That is, 1 s s 1 s 2 ε = sin A z cos A z = sin A z cos A z (radian) (7.3) 2 R R 2 R2 o since ∠Q = 90o , Therefore ∠ B = 90 − Az + ε

In spherical ΔPKB, side BK = ω (i.e. by construction) Using sine rule for spherical triangle;

K

sin ∠ s = sine of opposite sides

Q Az

s

Therefore, Each angle = its opposite side or its supplement . (180o − ∠) B (φ2, λ2) Let φ be the latitude of point B, then

A

(φ1, λ1) Fig 7.2 The geometry of the development of the Clarke’s Formula

PB = 90o − φ (co - latitude of point B)

∠PKB = 90o + φ ∠BKQ = 90o − φ




ω

Let

η be equal to spherical excess in ΔBQK

Therefore, ∠ B + ∠Q + ∠ K −η = 180o

∠Q = 90o Therefore, ∠B = 180o − 90o − (90o − φ ) +η But

K

In right angled triangle ΔABQ. QB = s sin A z

(7.5)

In right angled triangle ΔBQK.

Q A Az s (φ1, λ1)

(7.4)

B

tan( 90o − φ ) =

(φ2, λ2)


Thus QK =

QB

(7.5)

QK QB o

tan(90

− φ )

(7.6)

Substituting 7.5 into 7.6, we have

QK = s sin A z tanφ (7.7)



7.2 Clarke’s formula for Geodetic Line (Normal Section Formulae) Pole

Recall equation 7.3

P

ω

ε =

1 s 2 2 R2

then, K

A (φ1, λ1)

ε cos A z

=

1 s 2 2 R2

sin A z

(7.8)

1 BQ.QK Area Δ BQK 2 = The spherical excess η = (7.9) 2 R R2

Q Az

sin A z cos A z (radian)

B s


(φ2, λ2)

Substituting 7.5 and 7.7, i.e. for QB and QK into (7.9) we have

η =

1 s. sin A z s . .sin A z . tanφ 2 1 s 2 2 sin A z . tanφ (7.10) = 2 2 R 2 R



7.2 Clarke’s formula for Geodetic Line (Normal Section Formulae) K’

Considering the problem on a plane (Fig 7.3) and applying Legendre’s theorem Using sine law in ΔA’B’Q’, we can write

Q’

A' Q' 2

=

sin(90o − A z − ε ) 3

s ε sin(90o - ) 3

(7.11)

B’ s

Therefore, A' Q' =

A’

ε AQ ≡ A' Q' and since ε is small cos is close to 1 3 ⎛ 2 ⎞ Therefore, AQ ≡ s cos ⎜ A z − ε ⎟ (7.12) ⎝ 3 ⎠ And,

Fig 7.3 Geometry on a plane

2 ⎞ ⎛ cos ⎜ Az − ε ⎟ ε 3 ⎠ ⎝ cos 3 s




B' Q'

Similarly,

⎛ ⎝

ε ⎞ ⎛ ⎟ 3 ⎠ ⎝ ⎛ ε ⎞ s sin ⎜ A z − ⎟ ε ⎞ ⎝ 3 ⎠ = s sin ⎛ Therefore, B'Q' = ⎜ A z − ⎟ ε ⎞ ⎛ ⎝ 3 ⎠ sin ⎜ 90o − ⎟ 3 ⎠ ⎝ And, BQ ≡ B' Q' ⎛ ε ⎞ Therefore, BQ ≡ s sin ⎜ A z − ⎟ (7.13) ⎝ 3 ⎠ sin ⎜ A z −

Q’

B’ s

ε ⎞

s

=

⎟ 3 ⎠

sin ⎜ 90o −

A’ Also in triangle B'Q' K' ; Fig 7.3 Geometry on a plane

⎛ ⎝

BQsin ⎜ 90o − BK ≡ B' K ' =

η ⎞

⎟

3 ⎠ sin 90o − φ −η

(

)



7.2 Clarke’s formula for Geodetic Line (Normal Section Formulae) Substitute for BQ from 7.13 and η being small, i.e. η ⎞ ⎛ sin⎜ 90o − ⎟ ≡ 1 3 ⎠ ⎝

K’

Q’

Therefore,

⎛ η ⎞ ⎟ 3 ⎠ ⎝ BK = ⎛ η ⎞ cos ⎜φ − ⎟ ⎝ 3 ⎠ s sin ⎜ A z −

B’ s A’

But linear length of BK and in triangle BQK ϖ =

Fig 7.3 Geometry on a plane

BQ

⎛ ⎝

sin ⎜ 90o − φ −

η ⎞

⎟

3 ⎠

=

ϖ R;




Substitute for BQ from 7.13 we have

⎛ η ⎞ ⎟ R 3 ⎠ ⎝ ϖ = radians η ⎞ ⎛ cos ⎜φ − ⎟ ⎝ 3 ⎠ s

Q’

B’ s

In triangle B’Q’K’ and from sine law, we have QK =

A’ And again, Fig 7.3 Geometry on a plane

sin ⎜ A z −

Therefore,

⎛ 2 ⎞ .sin⎜φ + η ⎟ ⎛ o η ⎞ ⎝ 3 ⎠ sin⎜ 90 − ⎟ 3 ⎠ ⎝ ϖ R

η being small, i.e.

(

o

sin 90

⎛ 2 ⎞ QK = ϖ R.sin⎜φ + η ⎟ ⎝ 3 ⎠

−

)

η

≡

1

(7.14)



Spherical Triangle using Napier’s Rule (Normal Section Formulae) - Clarke Pole P

Considering Spherical Triangle BQK;

ω 90-K QK K

90-KB BQ 90-B

Q

A (φ1, λ1)

Az

B s


(φ2, λ2)

Sinus of an angle = Cosine of Opposites = Tangent Adjacents. e.g.

(

sin (QK ) = cos 90o − KB).cos (90o − B

= tan (90o − K ). tan ( BQ)

)



Solving the Spherical Triangle using Napier’s Rule (Normal Section Formulae) - Clarke Pole P

ω

In triangle BQK : sin φ = cos

QB

In triangle PBQ : sinφ = cos

QB

Therefore,

K

From fig. 7.1,

A (φ1, λ1)

Az

B s

(φ2, λ2)

R

(

. cos

QP R

= 90o − (φ +η ) QP R

is the co-latitude of point Q

Thus from which latitude of point AQ Q = (φ + η ) = φ Q Also φ Q = φ A + R Substitude for AQ from 7.12, we have;


)

QP R

Q

R

. cos 90o − (φ + η )

s ⎛ 2 ⎞ φ +η = φ A + . cos⎜ A z − ε ⎟ R ⎝ 3 ⎠



Solving the Spherical Triangle using Napier’s Rule (Normal Section Formulae) - Clarke Pole

Therefore,

ω

φ B K

⎛ 2 ⎞ . cos⎜ A z − ε ⎟ −η R ⎝ 3 ⎠

= φ A +

s

Where. H is the difference in latitude between B and Q. In similar manner, using Napier’s rule, in triangle BQK;

Q

A (φ1, λ1)

= φ B

Since as stated earlier,

P

(

)

sinψ = tan 90o − ω . tan

Az

B s

(φ2, λ2)

In triangle PBQ;

QK R

(

)

sinψ = tan 90o − ω . tan(90o − ∠ PBQ)

Therefore Fig 7.2 The geometry of the development of the Clarke’s Formula

QK R

= 90o − ∠PBQ

Thus

∠PBQ = 90o −

QK R



Solving the Spherical Triangle using Napier’s Rule (Normal Section Formulae) - Clarke Pole

ω

P

K

Substitute for QK from 7.14 we have

⎛ 2 ⎞ ω R . .sin ⎜φ + η ⎟ ⎝ 3 ⎠ = 90o − ω .sin (ψ + 2 η ) ∠ PBQ = 90o − R 3 In triangle ABQ of fig 7.1;

Q

Therefore

Az

B s

(φ2, λ2)

A (φ1, λ1) Fig 7.2 The geometry of the development of the Clarke’s Formula

∠Q = 90o

∠ ABQ = 90o − Az + ε

Thus, bearing BA (Back bearing of AB)

= 180o − (∠ PBQ + ∠ ABQ) ⎛ ⎛ o ⎞ 2 ⎞ ⎞ ⎛ o o ⎜ = 180 − ⎜ ⎜⎜ 90 − ω sin⎜ψ + η ⎟ ⎟⎟ + 90 − A z − ε ⎟⎟ ⎝ 3 ⎠ ⎠ ⎝ ⎝ ⎠ 2 ⎞ = A z + ω sin⎛ + ψ η ⎟ + ε (7.15) ⎜ ⎝ 3 ⎠



TRANSFORMATION TO THE ELLIPSOID (Normal Section Formulae) - Clarke ⎛ 2 ⎞ s. cos⎜ A z − ε ⎟ ⎝ 3 ⎠ −η ψ = φ B = φ A + M φ m ω

⎛ 1 ⎞ sin ⎜ Az − ε ⎟ s ⎝ 3 ⎠ radians : Difference in longitude = N . ⎛ η ⎞ φ B cos ⎜ψ + ⎟ ⎝ 3 ⎠

⎧ o ⎛ ψ + 2 η ⎞ − ε ⎫ = ± + 180 . sin B A ω ⎜ ⎟ ⎬ The azimuth of point B (i.e. azimuth BA) z ⎨ z ⎝ 3 ⎠ ⎭ ⎩ Where; M φ m

=

Radius of curvature in the meridian at mid-point φ m

N φ B

=

Radius of curvature in the prime vertical at point φ B



SUMMARY Direct problem using Clarke’s formula Given φ 1 , λ 1 ,α 12 , s Calculate

Must be given datum ellipsoid parameters

φ 2 , λ 2 ,α 21

Calculation steps:

A. Preliminary computations. a N = 1 1. Calculate N and M for point φ 1 2 2 (1 − e sin φ 1 ) 2

2. Calculate the 1st spherical excess

1 2 s sin A z . cos A z 2 ε = M φ 1 .N φ 1 . cosφ 1.(sin1" )

3. Calculate approximate change in Latitude;

4. Calculate approximate Latitude of point 2; 5. Calculate the 2nd spherical excess;

M =

Δφ =

s. cos A z M φ 1 .(sin1" )

ψ ' = φ 1 + Δφ

= ε tan A

tanψ '

a(1 − e 2 )

(1 − e

3

2

sin φ 1 )2 2

Where; A z = α 12



SUMMARY B. Precise computations. 1. Calculate the mean Latitude; 2. Calculate; N φ m

φ m

=

φ 1 + φ 2 2

and M φ m

3. Calculate 1st spherical excess; 4. Calculate the 2nd spherical excess;

1 2 s sin A z . cos A z 2 ε = M φ m . N φ m . cosφ m .(sin1")

= ε . tan A z . tanψ '

5. Calculate for ψ which is equal to latitude of point 2; i.e. φ 2

⎛ 2 ⎞ s. cos ⎜ A z − ε ⎟ ⎝ 3 ⎠ −η φ 2 = ψ = φ 1 + N φ m



SUMMARY C. Computations of Longitude of Point 2. 1. Calculate;

N φ 2

2. Calculate the difference in Longitude;

3. Longitude of point 2 ;

λ 2

⎛ 1 ⎞ sin ⎜ A z − ε ⎟ s ⎝ 3 ⎠ ω = . N φ B ⎛ 1 ⎞ cos ⎜φ 2 + η ⎟ ⎝ 3 ⎠

= 1 + ω

D. Computation of azimuth from Point 2 to 1 1. Azimuth at Point 2;

⎧ ⎩

⎛ + 2 η ⎞ − ε ⎫ ⎟ ⎬ ⎝ 3 ⎠ ⎭

B z = α 21 = α 12 ± ⎨180o + ω . sin⎜φ 2



7.3 CLARKE’S FORMULA FOR GEODETIC LINES (INVERSE PROBLEM) Given:

φ 1, λ 1,φ 2 , λ 2

α 12,α 21, s

Calculate;

From the solution of the direct problem, we have

⎛ 2 ⎞ s.cos ⎜ A z − ε ⎟ ⎝ 3 ⎠ −η (7.16) φ 2 = φ 1 +

And,

M φ m

⎛ 1 ⎞ sin ⎜ A z − ε ⎟ s ⎝ 3 ⎠ ω = . N φ m ⎛ 1 ⎞ cos ⎜ψ + η ⎟ ⎝ 3 ⎠

(7.17)

Rearranging 7.16 and 7.17, we have

⎛ 2 ⎞ M φ m (φ 2 −φ 1 +η ) = s.cos⎜ Az − ε ⎟ ⎝ 3 ⎠

(7.18)

⎛ 1 ⎞ ⎛ 1 ⎞ ω N . φ m . cos⎜ψ + η ⎟ = s.sin ⎜ Az − ε ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠

(7.19)

In above equations, all the parameters were known except the two spherical excess ε and η . Since they are small, we can compute the approximate values using s. cos A z = M φ

Δφ = M φ (φ 2 − φ 1 )

And,

1 ⎞ ⎛ s. sin A z = N φ cos φ m Δλ = s. sin ⎜ Az − ε ⎟

⎝

⎠



7.3 CLARKE’S FORMULA FOR GEODETIC LINES (INVERSE PROBLEM) Thus;

1 ε = 2

Ψ = φ 2

( s. cos A z )( s. sin A z )

η = ε . tan φ 2 .

M φ m . N φ m

Taking (7.19) / (7.18), we have; 1 ⎞ ⎛ s. sin ⎜ A z − ε ⎟ 3 ⎠ ⎝ ⎛ 2 ⎞ s. cos⎜ A z - ε ⎟ ⎝ 3 ⎠

Since ε is small;

⎛ 1 ⎞ N φ m . cos ⎜ψ + η ⎟ .ω ⎝ 3 ⎠ = (φ 2 − φ 1 + η ) M φ m

⎛ 1 ⎞ N φ m . cos ⎜ψ + η ⎟.ω ⎝ 3 ⎠ tan A z ' = (φ 2 − φ 1 + η ) M φ m

( s. sin A z ) = ε . tan A z . tan φ 2 ( s. cos A z )



7.3 CLARKE’S FORMULA FOR GEODETIC LINES (INVERSE PROBLEM) Therefore, the azimuth; 1 A z = α 12 = A z ' + ε .(1+ sin2 Az ' ) 3

And, azimuth;

(7.20)

⎛ 2 ⎞ B z = α21 = α12 ±180o + ω.sin⎜φ 2 + η⎟ − ε ⎝ 3 ⎠

From equation 7.18 we have;

The distance

s =

(φ 2 −φ 1 +η ) M . φ

m

⎛ 2 ⎞ cos⎜ A z − ε ⎟ ⎝ 3 ⎠

(7.22)

(7.21)



CALCULATION OF GEODETIC COORDINATES (Geodesic Formulae) Utilize the differential equations of the geodesic and are therefore intrinsically rigorous. The integration involves an elliptic integral. Such an integral can only be integrated in terms of elementary functions by expansion in series, after which integration is achieved term by term. Similarly, integration of any other differential equation of the geodesic of the ellipsoid will result in an infinite series. The series may be in terms of e 2 or higher order terms, or it may be in f and higher order terms. In both these cases, the accuracy of the formula will depend on the value of the truncated terms.



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Given the following information.

φ1

= 45o 20’ 30.2124”

λ1

= 8o 10’ 14.1247”

α12

= 300o 33’ 10.38”

s12 = 5321.732m. Calculate the value of φ2, λ2, α2 with reference to WGS84 reference ellipsoid using the GML method. Solution: For WGS84 ellipsoid.

a = 6378137m. 1/f = 298.257223563


SUG456 GEODESY : GEODETIC COORDINATES

7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Δλ Δφ

2

α21

The solution to this Direct Problem is

φ2 = ? = ?

φ1 = 45o 20’ 30.2124”

1

α12

φ = 0ο

300o 33’ 10.38”

λ2 = ?

λ1

= 8o 10’ 14.1247”

φ 2 = φ 1 + Δφ λ2 = λ1 + Δλ α21 = α12 −180o − Δα

Note that in GML formula, the mean elements are φ m = φ 1 + α

Δφ

= α12 +

2

Δα



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) First Iteration: We have

Δφ =

s.cosα m

M m.sin1"

And.

e2

Now,

1 f

= 2 f −

&

Δλ =

f 2

= 298.257223563

f =

1 298.257223563

= 1.124133936 ×10 −5 e2 = 0.00669438

f 2

s.sinα m N m. cosφ m.sin1"



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Calculate M m and N m using the related formula; M m

=

a (1 − e 2 )

(1 − e .sin 2

3

2

φ m )2

Where, we set φ m as 45o 20’ 30.2124” M m

= =

6378137(1 − 0.00669438)

(1 − 0.00669438 × sin 6335439.327 3

3

2

N m

=

2

=

(0.9966128841)2 =

6335439.327

0.9949236308 = 6,367,764.450m.

(1 − e .sin 2

)

45o 20' 30.2124"

a

=

1

2

φ m )2 6378137.000

(1 − 0.00669438 × sin 6378137.000 1

(0.9983050055)2 = 6,388,966.263m.

1

2

45 20' 30.2124")2 o



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Now,

Δφ =

s.cosα m

M m.sin1"

Where, we set α m as 300o 33’ 10.38” 5321 .732 × cos 300 o 33 ' 10 .38"

Δ φ =

6367764 .450 × sin 1"

=

5321 .732 × cos 300 o 33 ' 10 .38"

×

206265

6367764 . 450

= 87 .6275 ".

Δ λ =

s. sin α m N m . cos φ m . sin 1"

= =

5321 .732 × sin 300 o 33' 10 .38" × 206265 6388966 .263 × cos 45 o 20 ' 30 .2124 " − 210 .5003 "

Δ α = Δ λ . sin φ m = (− 210 .5003 ")× sin 45 o = −149 7313 "

20 ' 30 .2124 "



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Δφ

Then,

87.6275 "

= 43.8137 " 2 (− 210.5003 ") = = −105.2501" 2 2 Δα (−149.7313 ") = = −74.8656 " 2 2 2 Δλ

=

Hence, φ m

= φ 1 +

Δφ 2

= 45 o = 45 o

α m

= α 12 +

Δα 2

20 ' 30 .2124 " + 43 .8137 " 21' 14 .0261"

= 300 o 33 ' 10.38 " + (− 74 .8656 ") = 300 o 31'

55.5144 "



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Second Iteration: We have new values M m

=

(

a 1 − e2

(1 − e . sin 2

M m

=

φ m

= 45 o

21' 14 .03"

) 3

2

φ m

)

2

6378137(1 − 0.00669438)

(1 − 0.00669438 × sin

3

2

45 21' 14.03")2 o

= 6,367,778.080m. N m

=

a

(1 − e . sin 2

=

1

2

φ m

)

2

6378137.000

(1 − 0.00669438 × sin

= 6,388,970.820m.

1

2

45 21' 14.03")2 o

and

α m

= 300 o 3 1'

55.51 "



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Then,

Δφ =

Δφ =

s. cos α m M m . sin 1"

5321.732 × cos 300o 31' 55.51"

6367778.080 × sin 1" = 87.5733".

Δλ =

Δφ 2

=

87.5733" 2

= 43.7867"


=

5321.732 × sin 300 o 31' 55.51" × 206265

6388970.820 × cos 45o 21' 14.03" = −210.5903"

Δα = Δλ . sin φ m = (− 210.5903") × sin 45o 21' 14.03" = −149.8267"

Δλ (− 210.5903") = = −105.2952" 2

2

Δα (− 149.8267") = = −74.9134" 2

2



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Thus, φ m

α m

= φ 1 +

= α 12 +

Δφ 2

Δα 2

= 45 o

20 ' 30 .2124 " + 43 .7867 "

= 45 o

21' 13 .9991"

= 300 o 33 ' 10.38 " + (− 74 .9134 ") = 300 o 31'

55.47 "



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Third Iteration:

= 45 o 21' 14 .00" α m = 300 o 3 1' 55.47 " φ m

Now, we have new values

M m

=

(

(1 − e . sin 2

M m

=

)

a 1 − e2

3

2

φ m

)

2

6378137 (1 − 0.00669438 )

(1 − 0.00669438 × sin

3

2

)

o

45 21' 14.00"

2

= 6,367,778.070m. N m

=

a

(1 − e .sin 2

=

1

2

φ m

)

2

6378137.000

(1 − 0.00669438 × sin

= 6,388,970.817 m.

1

2

o

)

45 21' 14.00"

2



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Then,

Δφ =

s. cos α m M m . sin 1"

5321.732 × cos 300 o 31' 55.47"

Δφ =

6367778.070 × sin 1" = 87.5734".

Δλ =

Δφ 2

=

87.5734" 2

= 43.7867"


=

5321.732 × sin 300 o 31' 55.47" × 206265

6388970.817 × cos 45o 21' 14.00" = −210.5905"

Δα = Δλ . sin φ m = (− 210.5905") × sin 45o 21' 14.00" = −149.8269"

Δλ (− 210.5905") = = −105.2953" 2

2

Δα (− 149.8269") = = −74.9135" 2

2

Note: Iteration stopped when the difference in coordinates ( to previous iteration < 0.0001”.

,

,

)



7.3 GAUSS MID LATITUDE DIRECT PROBLEM (EXAMPLE.) Therefore, the values of the GML method;

φ2 λ2

= φ1 + Δφ

φ2, λ2, α2 with reference to WGS84 reference ellipsoid using

= 45o 20’ 30.2124” + 87.5734” = 45o 20’ 30.2124” + 0o 01’ 27.5734” =

= λ1 + Δλ = 8o 10’ 14.1247” + (- 210.5905”) = 8o 10’ 14.1247” - 0o 3’ 30.5905” =

α21

45o 21’ 57.7858”

= α12 + 180o +

Δα

8o 06’ 43.5342”

= 300o 33’ 10.38” - 180o + (-149.8269”) = 300o 33’ 10.38” - 180o - 0o 02’ 29.8269” = 120o 30’ 40.55”



7.3 GAUSS MID LATITUDE INVERSE PROBLEM (EXAMPLE.) Inverse Problem. Given, the geodetic coordinates of two points, namely 1 and 2. i.e. φ1, λ1, φ2, λ2 with reference to a reference ellipsoid. To find, azimuths α12, α21 and distance s12.

N m

=

a

(1 − e . sin 2

To find azimuth α12 ; tan α m

Where;

=

α 12

Δλ . N m . cos φ m Δφ M . m

Δφ = φ 2 − φ 1 Δλ = λ 2 − λ 1 φ + φ 2 φ m = 1 2

M m

=

(

a 1 − e2

(1 − e . sin 2

= α m −

) 3

2

φ m

)

2

Δα 2

1

2

φ m )2

Δα = Δλ . sin φ m α 21

= α 12 + Δα ± 180 o

s12

=

Δφ M . m . sin 1" cos α m

No iteration is needed for the above problem where mean value of latitude can be directly determined with the coordinates of point 1 and 2 known.

SUG456-4N.pdf

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