Fundamental Parameters of The Ellipsoid, The Meridian Ellipse, and Coordinate Conversion
Fundamental Parameters of The Ellipsoid, The Meridian Ellipse, and Coordinate Conversion
Fundamental Parameters of The Ellipsoid, The Meridian Ellipse, and Coordinate Conversion
Fundamental
Fundamental Parameters
Parameters ofof the
the
Ellipsoid,
Ellipsoid, the
the Meridian
Meridian Ellipse,
Ellipse, and
and
Coordinate
Coordinate Conversion
Conversion
Lecture No. 7
a.s. caparas/06
The fundamental
parameters of the ellipse
Formulas:
are:
a−b
1. Flattening or Polar f =
a
Flattening, f
a 2 -b 2 a 2 − b2
e= ; e2 =
2. First Eccentricity, e a a2
a 2 -b 2 a 2 − b2
3. Second Eccentricity, e’ e' =
b
; (e')2 =
b2
4. Angular Eccentricity, α cos α = 1 − f ; sin α = e ; tan α = e'
1
Latitudes on the Meridian Ellipse
There are three different
latitudes used to define the
position of the point on a
meridian ellipse: z
1.Geodetic Latitude (φ)- angle
between the line normal to the
point and the equatorial plane.
2.Geocentric Latitude (ψ)- angle P’
between the line connecting the p
center of the ellipse to the point P
and the equatorial plane. a r
b
3.Reduced Latitude(β)- obtained z
by projecting the ellipse on the β ψ φ
geocentric circle having a radius p
a
equal to the semi-major axis, a
2
Relationship Between the Various
Latitude
Comparing the parametric representations of
the meridian ellipse using the different latitudes,
we can find transformation between φ, β, and ψ:
• Geocentric to Geodetic:
2
b
tan ψ = tan ϕ
a
• Reduced to Geodetic:
b
tan β = tan ϕ
a
3
Example Problem
Problem: therefore:
A point on the ellipsoid has a a
2
x=(p+hcos φ)cos λ
y =(p+hcos φ)sin λ
z=(z+hsin φ)
4
Example Problem
Problem: Using the equations for converting
A point on the ellipsoid has a geodetic to cartesian:
geodetic coordinates φ=45°N, λ x=(p+hcos φ)cos λ
=121°E, and h=1500 m. If the y =(p+hcos φ)sin λ
flattening f of the ellipsoid is z=(z+hsin φ)
1/294.98 and the semi-major axis
a=6,378,206 m, compute the Solving for p and z:
space rectangular coordinates of a cos ϕ a(1 − e 2 ) sin ϕ
the points p= ,z=
1 − e 2 sin 2 ϕ 1 − e 2 sin 2 ϕ
Example Problem
Solving for z: z=(4487145.279+1500sin 45)
z=4,488,205.939 m
6378206.4(1− 0.00676865799760962)sin 45o
z=
1− (0.00676865799760962)sin2 45o
The space rectangular
z = 4,487,145.279 m coordinates of the point are:
y =(4517724.209+1500cos 45)sin
121
y =3,873,354.629 m
5
Geodetic Coordinates and the Space Rectangular
Coordinates
• We can get the geodeteic
coordinates (φ, λ, h) of a point
given its space rectangular
coordinates (x,y,z) using these
equations:
• However, most of the solution
in converting space
rectangular coordinates to
geodtic coordinates requires
iteration in the computation of
the geodetic latitude.
• There are several solutions
that can be used in this
conversion
z N
−1
ϕo = tan−1
z
ϕ = tan −1
1− e
2
x2 + y 2 x2 + y2 N + h
6
Geodetic Coordinates and the Space Rectangular
Coordinates
Another solution: 3.Then compute for h:
1.Calculate
y x2 + y 2
λ = tan−1 h= −N
x cos ϕ
2.Iterate for φ using as an
initial value for φ: where:
z e2 N sin ϕ a
ϕ = tan−1 1+ N=
x2 + y 2 z (1 − e sin 2 ϕ )1/ 2
2
z
ϕinitial = tan−1
(1− e ) x + y
2 2 2
y
λ = tan−1 in which:
x
p = x2 + y 2
z + e a sin µ
2 3
ϕ = tan−1
p − e a cos µ
2 3 r = p2 + z 2
a2 z(1− f ) ae2
h = p cosϕ + z sin ϕ − tan µ = 1+
p r
N