Lecture 8

GE 161 – Geomet Geometric ric Geodesy Geodesy The Reference Ellipsoid and the Computation of the Geodetic Posi Posi tion: tion: Properties of the Ellipsoid

Radii Radii of of Curvature Curvature on on the the Ellipsoid Ellipsoid and and Radii Radii of Spherical Approximation Approximation of of the the Earth Earth

Lecture No. 8 Department of Geodetic Engineering University of the Philippines a.s. caparas/06

Normal Sections on the Ellipsoid • •

•

•

•

Lecture 8

Consid Cons ider er fi firs rstt a no norm rmal al to th the e surface of the ellipsoid at some point. A parti particu cula larr pla plane ne wil willl cut cut the the surface of the ellipsoid forming a curve which is known as the normal section. At ea each ch po poin intt the there re ex exis istt an an infinite number of normal section as there exist an infinite number planes that that contain the normal line. Howe Ho weve ver, r, at ea each ch po poin int, t, th ther ere e exist two mutually perpendicular normal sections whose curvature will be maximum and minimum. Thes Th ese e nor norma mall sec secti tion ons s is is called the principal normal sections. GE 161 161 – – Geometric Geodesy

The Reference Ellipsoid and the Computation of the Geodetic Position: Properties of the Ellipsoid

Principal Normal Sections On the ellipsoid the two principal normal sections are:

Meridional Normal Section

1.The Meridian or Meridional Merid ional Norm Normal al Section – Section – a plane plane passi passing ng through the point and the two poles. 2.The Prime Vertical Normal Section – Section – a pla plane ne passing through the point and perpendicular to the meridian at that point. Lecture 8

Prime Vertical Normal Section

GE 161 161 – – Geometric Geodesy


Radii of Curvature of a Normal Sections In order to find the radius of curvature of any normal section at any arbitrary direction, we may utilize the Euler’s formula:

1 ρ

=

2 2 cos θ sin θ

ρ1

+

ρ2

where: ρ= is the radius of curvature of the section (any arbitrary section) θ = is the angle measured from the meridian of the point curvature of the principal principal normal section with the ρ1=is the radius of curvature maximum curvature curvature of the principal principal normal section with the ρ2=is the radius of curvature manimum manim um curva curvature ture

Lecture 8



Radius of Curvature of the Principal Normal Sections • Merid Meridional ional Radiu Radius s of Curvatu Curvature, re, M:

a (1 − e2 )

M=

3

(1 − e sin 2

2

ϕ) 2

at the equator: Mϕ=0 = a(1 − e2 ) = a(1− f )2

at the poles: M ϕ=90

=

a (1 − e 2 ) 3

(1 − e 2 ) 2

Lecture 8

=

a 1

=

(1 − e 2 ) 2

a (1 − f )



Radius of Curvature of the Principal Normal Sections • If th the e val value ues s of of th the eM were tabulated, they could be plotted with respect to an origin at the surface of the reference ellipsoid. • The The en endp dpoi oint nts s of th the e various M values would fall on a curve known as the locus of the centers of the merid meridional ional radiu radius s of curvature. Lecture 8

∆2


∆1

locus of the centers of the curvature of the meridian


Radius of Curvature of the Principal Normal Sections • Prime Vertical Radius of Curvature p=Ncosφ

N =

a 1

(1 − e sin 2

2

φ

ϕ) 2

At the equator: Nφ=0=a At the poles: N ϕ=90 Lecture 8

=

a

p

(1 − f ) GE 161 161 – – Geometric Geodesy


Comparing M and N… • We can can see see that that M and and N are are minim minimum um at at points on the equator. • At the the poles poles M and and N are are equal equal wit with h value value equal to a/(1-f). • If we tak take e the the ra ratio tion n of M and N, we wil willl find that: N (1 − e2 sin2 ϕ) = 2 M (1 − e ) • Thus, N≥M where equality holds at the poles. Lecture 8



Radius of Curvature of the normal section at any given azimuth • using using Euler Euler’s ’s formu formula la we can can deter determin mine e the radius of curvature letting θ =α =azimuth of the normal section from the north, ρ1= N and ρ2=M by: 1 R α R α

Lecture 8

= =

sin2 α cos2 α N

+

M

MN N cos2 α + M sin2 α GE 161 161 – – Geometric Geodesy


Gaussian Mean Radius of Curvature of Normal Sections at a point • The Gau Gaussi ssian an Mean Mean Rad Radiu ius s R of all the radii of curvature of all the normal section containing the normal line is given by:

R = MN • The valu value e of R is help helpful ful when when a radius radius of of a sphere that is to approximate the ellipsoid is required. Lecture 8



Example Problem Solving for N: N =

Problem: Compute for the radii of curvature of the two principal normal sections and the Gaussian Mean radius of curvature at the point whose geodetic latitude is 45 °N on the Clarke Spheroid of 1866.

N =

a 1

(1 − e 2 sin 2 ϕ) 2 6378206 1

(1 − 0.006768628177 sin 2 45o ) 2

N = 6,389,026.399 m Solving for M:

Solution: Given: φ=45°N f=1/294.98 a=6,378,206 m e2=0.006768628177 Find: N, M, and R

a (1

M

3

(1

M

2

e )

=

2

e sin

2

)2

6378206 (1 − 0.0067686281 77 ) 3

(1 − 0.0067862817 7 sin 2 45 o ) 2

M = 6,367,330.501 m Solving for R: R = MN = (6367330.501)(6389026.399

R = 6,378,169.225 m Lecture 8



Earth as a Sphere • Since Since the the comput computati ation on of som some e quanti quantitie ties s on the the surface of the ellipsoid is sometimes too complex to handle, geodesists uses the sphere as a model. • This This reduces reduces the the complex complexity ity of of derivi deriving ng formu formulas las and evaluating quantities. • In orde orderr for for us to use use a spher sphere e as a referen reference ce model, we need to find a sphere which is equivalent to the reference ellipsoid that we are using Lecture 8



Earth as a Sphere • There There are are severa severall way way of find finding ing a sphere sphere equivalent to the reference ellipsoid: 1. Equal surface area 2. Equal volume 3. Ellipsoid’s mean radius - Ga Gaus ussi sian an - Mean of the the three three semi-a semi-axes xes

Lecture 8



Radii Approximation to the Earth or Mean Radius of the Earth as a Sphere

A suitable radius may be defined by equating the expressions of the quantities being compared: 1.

Sphe Sp heri rica call radiu radius s havi having ng the the sam same e area area s th the e ellip ellipso soid id

17 4 67 6   1 R A = a 1 − e2 − e − e .... 360 3024  6  2.

Spheri Sphe rica call radi radius us hav havin ing g the the same same Vol Volum ume e as the the ellipsoid

R v Lecture 8

=

3

2

ab



Radii Approximation to the Earth or Mean Radius of the Earth as a Sphere

3. Sphe Spheri rica call radi radius us hav havin ing g the the mean mean rad radiu ius s of the three semi-axes of the ellipsoid

R m =

(a + a + b) 3

4. Gaus Gaussi sian an mea mean n rad radiu ius s as th the e radi radius us of of the the sphere

R= MN Lecture 8



Example Problem Problem: What are the radii of the equivalent spheres of the Clarke Spheroid of 1866. Solution: Given: f=1/294.98

a=6,378,206 m e2=0.006768628177 Find: Rm, RA, and RV

Lecture 8

Solving for Rm: (a + a + b)

R m

=

R m

=

R m

= 6,370,998.499 m

3 (6378206 + 6378206 + 6356583.497) 3

Solving for RA: 17 4 67 6   1 R A = a 1 − e 2 − e − e .... 360 3024  6  R A

= 6,370,996.873

m

Solving for Rv: R v

= 3 a 2 b = 3 (6378206) 2 (6356583.497)

R v

= 6,370,990.339 m



Lecture 8

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