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The Direct Geodetic Problem Using the Gauss Mid-Latitude Method | SURE 452, Study notes of Engineering

Material Type: Notes; Class: Geodesy 1; Subject: Surveying Engineering; University: Ferris State University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/07/2009

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Gauss Mid-Latitude Direct Page 1 of 5
The radius of curvature in the meridian, M, and Prime Vertical, N, for the first station:
The first, or initial, iteration
Direct Problem:
---------------------------------------------------------------------------------------------------------------------------------------------
α12 radians az()
:=
az 127.1027080:=s 54972.161m:=
λ1l radians λ
()
:=φ1p radians φ
()
:=
lifλ0<1, 1,
()
:=pifφ0<1, 1,
()
:=
λ43.55306630:=φ 37.39155571:=
Given data:
---------------------------------------------------------------------------------------------------------------------------------------------
e20.00669454=
e2a2b2
a2
:=
first eccentricity squared:
b 6356774.71930860 m=
baaf:=f1
298.25000158005
:=a 6378160 m:=
The following data refer to a given reference system
---------------------------------------------------------------------------------------------------------------------------------------------
r2d 180.0
π
:=
dms ang( ) degree floor ang()
rem ang degree()60
mins floor rem()
rem1 rem mins()
secs rem1 60.0
degree mins
100
+secs
10000
+
:=
radians ang( ) d dd ang()
dπ
180.0
:=dd ang( ) degree floor ang 0.0000000001+()
mins ang degree( ) 100.0
minutes floor mins 0.000000001+()
seconds mins minutes( ) 100.0
degree minutes
60.0
+seconds
3600.0
+
:=
Some useful angle functions:
The Direct Geodetic Problem Using the Gauss Mid-Latitude
Method
pf3
pf4
pf5

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The radius of curvature in the meridian, M, and Prime Vertical, N, for the first station:

The first, or initial, iteration

Direct Problem:

α 12

:=radians az( )

s :=54972.161m az :=127.

λ 1

φ :=l radians⋅ (^ λ)

:=p radians⋅ (^ φ )

p :=if (^ φ < 0 , − 1 , 1 ) l :=if (^ λ < 0 , − 1 , 1 )

Given data: φ :=−37.39155571 λ :=43.

e 2

e 2

a

2 b

2 −

a

2

first eccentricity squared: :=

b =6356774.71930860 m

f b :=a −a f⋅

a :=6378160 m⋅ :=

The following data refer to a given reference system

r2d

π

dms ang( ) degree ←floor ang( )

rem ←(ang −degree) 60⋅

mins ←floor rem( )

rem1 ←( rem −mins)

secs ←rem1 60.0⋅

degree

mins

secs

radians ang( ) d ←dd ang( )

d

π

dd ang( ) degree ←floor ang( +0.0000000001) :=

mins ←(ang −degree) 100.0⋅

minutes ←floor mins( +0.000000001)

seconds ←(mins −minutes) 100.0⋅

degree

minutes

seconds

Some useful angle functions:

The Direct Geodetic Problem Using the Gauss Mid-Latitude

Method

N

m

=6386196.61404326 m

N

m

a

1 e 2

sin φ

( ( m))

2

− ⋅

M

m

=6359439.76713680 m

M

m

a 1 e 2

1 e 2

sin φ m

2

− ⋅

Compute the radius of curvature in the meridian and prime vertical based on the mean latitude

φ The estimate of the mean latitude m

φ 1

φ 2

λ 2

λ 1

φ := +∆λ 2

φ 1

:= +∆φ

Compute the latitude and longitude of the second point

Begin the second iteration towards the solution

The subscript b indicates that the changes are the

base values for the change in longitude and latitude to

be used to see when the iteration stops.

∆φ b

∆λ :=∆φ b

:=∆λ

p dms⋅ (^ ∆φ ⋅r2d)=−0.

∆φ p :=if (^ ∆φ < 0 , − 1 , 1 )

s cos α

M cos

∆λ

l dms⋅ (^ ∆λ ⋅r2d)=0.

∆λ l :=if (^ ∆λ < 0 , − 1 , 1 )

s sin α

N cos φ 1

Compute an initial approximation of the change in latitude, ∆φ, and change in longitude, ∆λ:

N =6386142.43899800 m

N

a

1 e 2

sin φ

2

− ⋅

M =6359277.92432075 m

M

a 1 e 2

1 e 2

sin φ 1

2

− ⋅

N

m

=6386196.42190237 m

N

m

a

1 e 2

sin φ

( ( m))

2

− ⋅

M

m

=6359439.19312923 m

M

m

a 1 e 2

1 e 2

sin φ m

2

− ⋅

φ m

φ 1

φ 2

λ 2

λ 1

φ := +∆λ 2

φ 1

:= +∆φ

Begin the fourth iteration:

∆φ b

∆λ :=∆φ b

:=∆λ

dms ∆φ ∆φ b

− ⋅r2d

Since the difference is still greater

than 0.01", another iteration will be

run.

dms ∆λ ∆λ b

− ⋅r2d

p dms⋅ (^ ∆φ ⋅r2d)=−0.

∆φ

s cos α 12

∆α

M

m

cos

∆λ

p :=if (^ ∆φ < 0 , − 1 , 1 )

l dms⋅ (^ ∆λ ⋅r2d)=0.

l :=if (^ ∆λ < 0 , − 1 , 1 )

∆λ

s sin α 12

∆α

N

m

cos φ m

dms (^ ∆α ⋅r2d)^ =−0.

∆α ∆λ

sin φ

( m)

cos

∆φ

∆λ

3

sin φ

( m)

cos

∆φ

cos φ m

3

cos

∆φ

3

α 21

α =306. 21

dms α 12

  • ∆α+i

⋅r2d

i if α 12

+∆α

< π, π,−π

The back azimuth from 2 to 1 is:

λ 2

φ =44. 2

The latitude of point 2 is: The longitude of point 2 is:

λ 2

l dms λ 2

⋅r2d

φ := ⋅ 2

p dms φ 2

⋅r2d

l if λ 2

p if φ := 2

λ 2

λ 1

φ := +∆λ 2

φ 1

:= +∆φ

The latitude and longitude of the second point are:

dms ∆φ ∆φ b

− ⋅r2d

dms ∆λ ∆λ^ Stop the calculations. b

− ⋅r2d

p dms⋅ (^ ∆φ ⋅r2d)=−0.

∆φ

s cos α 12

∆α

M

m

cos

∆λ

p :=if (^ ∆φ < 0 , − 1 , 1 )

l dms⋅ (^ ∆λ ⋅r2d)=0.

l :=if (^ ∆λ < 0 , − 1 , 1 )

∆λ

s sin α 12

∆α

N

m

cos φ m

dms (^ ∆α ⋅r2d)^ =−0.

∆α ∆λ

sin φ

( m)

cos

∆φ

∆λ

3

sin φ

( m)

cos

∆φ

cos φ m

3

cos

∆φ

3