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Material Type: Notes; Class: Geodesy 1; Subject: Surveying Engineering; University: Ferris State University; Term: Unknown 1989;
Typology: Study notes
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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
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
m
=6386196.61404326 m
m
a
1 e 2
sin φ
2
− ⋅
m
=6359439.76713680 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
s cos α
M cos
∆λ
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
a
1 e 2
sin φ
2
− ⋅
M =6359277.92432075 m
a 1 e 2
1 e 2
sin φ 1
2
− ⋅
m
=6386196.42190237 m
m
a
1 e 2
sin φ
2
− ⋅
m
=6359439.19312923 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
∆φ
s cos α 12
∆α
m
cos
∆λ
∆λ
s sin α 12
∆α
m
cos φ m
∆α ∆λ
sin φ
cos
∆φ
∆λ
3
sin φ
cos
∆φ
cos φ m
3
cos
∆φ
3
α 21
α =306. 21
dms α 12
⋅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
∆φ
s cos α 12
∆α
m
cos
∆λ
∆λ
s sin α 12
∆α
m
cos φ m
∆α ∆λ
sin φ
cos
∆φ
∆λ
3
sin φ
cos
∆φ
cos φ m
3
cos
∆φ
3