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DIRECT AND INVERSE GEODETIC PROBLEM
Surveying Engineering Department
Ferris State University
The direct and inverse geodetic problems are one of the most basic questions posed to the geodesist/surveyor. The direct problem can be simply stated as given the latitude and longitude of a beginning point and a distance and azimuth to the second point, compute the latitude and longitude of that second point along with the back azimuth from point 2 to 1. The inverse problem can be formulated as given the latitude and longitude of two points, compute the distance between them along with the forward and reverse azimuths. The problem is complex because the earth is not a plane, or even a sphere. Thus, to solve these problems will require assumptions that limit the accuracy of the results.
Bowring Formulas
Direct Problem
Bowring developed a formulation for the direct problem using a conformal projection of the ellipsoid on the sphere and it is called a Gaussian projection of a second kind. The simplicity of the system lies in that the ellipsoidal geodesic is projected to its corresponding line on a sphere thereby allowing the formulation using spherical trigonometry. Both the direct and inverse solutions are non-iterative. These formulas are valid for lines up to 150 km.
Without derivation, the common equations used by Bowring are:
( )
( )
( )
A e
B e
C e
w
A
2 4 1
2 2 1
2
2 1
' cos
' cos
ϕ
ϕ
λ λ
Then the formulas for the direct problem are presented as follows:
σ
λ λ
σ α ϕ σ ϕ α
σ α ϕ α
ϕ ϕ ϕ
α
α σ σ ϕ α
^
−
−
−
s B a C
A
A
B
D
A
w
D B e D BD
B
B
2
2 1
(^1 ) 1 1 12
1 12 1 12
2 1
2 1
12
(^1 ) 1 12
tan
tan sin cos tan sin cos
sin sin cos sin sin tan
' sin
tan
sin cos tan tan cos
Inverse Problem
The inverse problem begins by computing a series of constant values.
sinw (B cos cosD sin sinD)
A
F
E sinDcosw
sin 2 4 B
3 e' 1 2 B
D
1 1
2 1
2
= ϕ − ϕ
∆ϕ
( )
= ϕ + ϕ
σ = +
sin Bcos tanDtan w A
tanH
sin 2 E F
E
tanG F
1 1
(^2212)
- Solve for the approximate change in latitude (∆ϕ) by using the measured azimuth
instead of (^) ( α 12 + ∆α^2 ) and using the radius of curvature in the meridian at point 1 (M 1 ) instead of the mean radius of curvature (Mm).
- Compute the first approximation of the latitude of the second point
ϕ 2 = ϕ 1 + ∆ϕ
- Determine the first approximation of the change in longitude and the longitude of the second point.
λ 2 = λ 1 + ∆λ
- Find the first approximation of the change in azimuth.
- Using these approximations, update the values M, N, ϕm and the other values listed in the first four steps.
Inverse Problem
The Gauss Mid-Latitude formulae for the inverse problem can be developed into direct computation requiring no iteration. It can be presented as follows (without derivation)
s s
s N s N X X m
m
o
−
1
1
1
12
(^1 ) 2
21 12
sin
α tan
α α
∆α
∆α
where: s 1 ( X 12 X 22 )
(^12) = +. Then,
X s N
X s M
F
F
m m
m
m
m m
1 1 12
2 1 12
3
2
=^
sin 'cos
cos 'cos
sin
sin
sin sec
sin cos
α ϕ
α
ϕ
ϕ ϕ
∆α ∆λ
∆α ∆ϕ
∆λ
∆ϕ ∆ϕ
∆ϕ
∆ϕ
∆λ ∆λ
∆λ ∆λ
∆α ∆λ
∆ϕ ∆λ
This approach has an accuracy of about 1 ppm for lines up to 100 km in length.
PUISSANT FORMULAS
The Puissant formulas are not generally used for lines greater than 100 km long. The formulation of this method is based on defining spheres passing through the first point.
Direct Problem
Without derivation, the difference in the latitude can be presented as:
∆ϕ = s cos α 12 B − s sin α C − h s sin α E − ( δϕ)D
2 2 12
2 2 12
2
where:
For short lines up to about 19 km, the Puissant equations can be simplified into the following form.
m
12 2 2
2 12
2 2 12
sin
sin sec N
s
scos B s sin C D
∆α=∆λ ϕ
∆λ= α ϕ
∆ϕ= α − α −δϕ
Inverse Problem
The inverse problem is an iterative solution. First, compute
2 (^2122 m) 2
2
2 2 12 1 sin sec 6 N
s 1
N cos ssin − − α ϕ
∆λ ϕ α =
For the first initial approximtion, the numerator is set to 1. Then, solve for
[ ssin C hEs sin ( )D] B
s cos 2 12
2 2 12
2 2 α 12 = ∆ϕ+ α + α + δϕ
Even though h is not known on the right hand side of the equation, one can obtain an estimate for s cosα 12. The value for h can be computed in subsequent iterations of these
equations. Next, solve for α 12 using
12
12 (^12) scos
ssin tan α
α α =
then
[ ( ) ( )] 2 21 12
2 s = ssinα 12 +scosα
Iterate to a solution and solve for the back azimuth using the same relationship given in the solution for the direct problem.
