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

Ferris State University (FSU)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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koofers-user-6n4 🇺🇸

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bg1
φA37.39155571−:= λA43.55306630:=
pifφA0<1−, 1,
()
:= lifλA0<1−, 1,
()
:=
φ1p radians φA
()
⋅:= λ1l radians λA
()
⋅:=
φB37.570912874−:= λB44.252481672:=
pifφB0<1−, 1,
()
:= lifλB0<1−, 1,
()
:=
φ2p radians φB
()
⋅:= λ2l radians λB
()
⋅:=
-----------------------------------------------------------------------------------------------------------------------------------------
-
Inverse Problem:
The mean latitude:
φm
φ1φ2
+
2
:= φmr2d⋅37.80342859−=
The Inverse Geodetic Problem Using the Gauss Mid-Latitude
Method
Some useful angle functions:
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
+
:= radians ang( ) d dd ang()←
dπ
180.0
⋅
:=
dms ang( ) degree floor ang()←
rem ang degree−()60⋅←
mins floor rem()←
rem1 rem mins−()←
secs rem1 60.0⋅←
degree mins
100
+secs
10000
+
:=
r2d 180.0
π
:=
-----------------------------------------------------------------------------------------------------------------------------------------
-
The following data refer to a given reference system
a 6378160 m⋅:= f1
298.257222028
:= baaf⋅−:=
b 6356775.23702048 m=
first eccentricity squared: e2a2b2
−
a2
:= e20.006694380024537=
-----------------------------------------------------------------------------------------------------------------------------------------
-
Given data:
pf3

Partial preview of the text

Download The Inverse geodetic Problem Using the Gauss Mid-Latitude Method | SURE 452 and more Study notes Engineering in PDF only on Docsity!

φ A

:= −37.39155571 λ A

p if φ A

:= l if λ A

φ 1

p radians φ A

:= ⋅ λ 1

l radians λ A

φ B

:= −37.570912874 λ B

p if φ B

:= l if λ B

φ 2

p radians φ B

:= ⋅ λ 2

l radians λ B

Inverse Problem:

The mean latitude:

φ m

φ 1

φ 2

:= φ m

⋅r2d =−37.

The Inverse Geodetic Problem Using the Gauss Mid-Latitude

Method

Some useful angle functions:

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

:= radians ang( ) d ←dd ang( )

d

π

dms ang( ) degree ←floor ang( )

rem ←(ang −degree) 60⋅

mins ←floor rem( )

rem1 ←( rem −mins)

secs ←rem1 60.0⋅

degree

mins

secs

r2d

π

The following data refer to a given reference system

a := 6378160 m⋅ f

:= b :=a −a f⋅

b =6356775.23702048 m

first eccentricity squared: (^) e 2

a

2 b

2 −

a

2

e 2

Given data:

Gauss Mid-Latitude Method - Inverse

Problem

Page 2 of 3

s i

s =54971.99248763 m i

X

2 X 2

2 := +

X

X =−33099.40217128 m 2

∆φ p

cos

∆λ

B

m

X

X =43890.19860931 m 1

∆λ p

cos φ m

A

m

∆λ p

∆λ ⋅ r2d=0. p

∆λ

sin

∆λ

∆λ

2

∆φ p

∆φ ⋅r2d =−0. p

∆φ

sin

∆φ

∆φ

2

∆A ∆λ dms (∆^ A r2d⋅ )^ =−0.

sin φ m

cos

∆φ

F ∆λ

3

  • ⋅

F F =−0.

sin φ

( m)

⋅ cos φ

( ( m))

2 := ⋅

B

m

m

B =

m

M

m

M

m

M =6359439.64734370 m m

a 1 e 2

W

m

3

A

m

m

A =

m

N

m

N

m

N =6386196.22720554 m m

a

W

m

W

m

W =0.

m

1 e 2

sin φ

( ( m))

2 := − ⋅

∆φ φ ∆φ ⋅ r2d=−0. 2

φ 1

∆λ λ ∆λ ⋅ r2d=0. 2

λ 1

The difference in longitude and latitude: