Map_projections_II

Map projections

Author: Dr.-Ing. Norbert R¨osch Karlsruhe Institute of Technology (KIT) Geodetic Institute Karlsruhe Englerstraße 7 76128 Karlsruhe Second draft (March 2011)

Contents 1 Introduction

1

1.1

What are map projections good for? . . . . . . . . . . . . . .

1

1.2

Classification . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2 Fundamentals

2

2.1

The Gaussian curvilinear parameters u and v

. . . . . . . . .

2

2.2

The line element . . . . . . . . . . . . . . . . . . . . . . . . .

3

2.3

Intersection angles of surface curves . . . . . . . . . . . . . . .

8

2.4

The infinitesimal area . . . . . . . . . . . . . . . . . . . . . . .

9

3 Theory of Distortions

11

3.1

The scale distortion . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2

The distortion of areas . . . . . . . . . . . . . . . . . . . . . . 17

3.3

The distortion of the azimuth . . . . . . . . . . . . . . . . . . 18

3.4

The infinitesimal angular distortion . . . . . . . . . . . . . . . 19

3.5

The maximal distortion of the azimuth . . . . . . . . . . . . . 21

3.6

Ellipse of distortion . . . . . . . . . . . . . . . . . . . . . . . . 21

3.7

Projections with specific properties . . . . . . . . . . . . . . . 24 3.7.1

Equidistant projections . . . . . . . . . . . . . . . . . . 24

3.7.2

Equal area projections . . . . . . . . . . . . . . . . . . 27

3.7.3

Conformal projections . . . . . . . . . . . . . . . . . . 28 i

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Script ”Map Projections”

4 Oblique map projections

29

4.1

From Gaussian parameters to geographic coordinates . . . . . 29

4.2

The azimuthal coordinates . . . . . . . . . . . . . . . . . . . . 30

4.3

Specific curves on the datum surface . . . . . . . . . . . . . . 31 4.3.1

The loxodrome . . . . . . . . . . . . . . . . . . . . . . 31

4.3.2

The orthodrome . . . . . . . . . . . . . . . . . . . . . . 38

5 Map projections with an orthogonal parameter system 5.1

5.2

5.3

40

Conic projections . . . . . . . . . . . . . . . . . . . . . . . . . 40 5.1.1

Equal area conic projections . . . . . . . . . . . . . . . 41

5.1.2

Conformal conic projections . . . . . . . . . . . . . . . 45

Cylinder projections . . . . . . . . . . . . . . . . . . . . . . . 47 5.2.1

Equal area cylinder projections . . . . . . . . . . . . . 48

5.2.2

Conformal cylinder projections . . . . . . . . . . . . . 48

Azimuthal projections . . . . . . . . . . . . . . . . . . . . . . 49 5.3.1

Equal area azimuthal projection . . . . . . . . . . . . . 50

5.3.2

Conformal azimuthal projection . . . . . . . . . . . . . 51

5.3.3

The gonomonic projcection . . . . . . . . . . . . . . . . 52

5.3.4

The orthographic projection . . . . . . . . . . . . . . . 52

6 Generalized projections

54

6.1

Pseudoconic projections . . . . . . . . . . . . . . . . . . . . . 54

6.2

Pseudocylindrical projections . . . . . . . . . . . . . . . . . . 59 6.2.1

Equal area with equidistant parallels . . . . . . . . . . 61

6.2.2

Equal area with elliptical meridians . . . . . . . . . . . 62

6.2.3

Equal area with elliptical meridians and pole lines . . . 65

6.2.4

With equidistant equator and equidistant central meridian 65

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch 7 Special projections and quality aspects 7.1

7.2

iii 69

Aphylactic small scale maps . . . . . . . . . . . . . . . . . . . 70 7.1.1

The Robinson projection . . . . . . . . . . . . . . . . . 70

7.1.2

Chamberlins trimetric projection . . . . . . . . . . . . 76

Optimal projections . . . . . . . . . . . . . . . . . . . . . . . . 77 7.2.1

Local distortion quantities . . . . . . . . . . . . . . . . 78 7.2.1.1

7.2.2

The image of the finite circle . . . . . . . . . 79

Global distortion quantities . . . . . . . . . . . . . . . 80 7.2.2.1

The concept of Peters . . . . . . . . . . . . . 80

7.2.2.2

Canters’ scattered circles . . . . . . . . . . . . 82

7.2.2.3

Capek and the limits . . . . . . . . . . . . . . 84

7.2.2.4

Approaches based on the distortion ellipse . . 85

8 Further concepts – isometric coordinates

86

8.1

Isometric surface parameters . . . . . . . . . . . . . . . . . . . 86

8.2

Isometric parameters on the sphere . . . . . . . . . . . . . . . 88

8.3

Conformal projection of the sphere . . . . . . . . . . . . . . . 89 8.3.1

Mercator projection . . . . . . . . . . . . . . . . . . . . 89

8.3.2

Stereographic projection . . . . . . . . . . . . . . . . . 90

8.3.3

Projection of the sphere on a sector . . . . . . . . . . . 91

A Exercises

93

B Mathematical explanations

103

B.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 B.2 Spherical trigonometry . . . . . . . . . . . . . . . . . . . . . . 104 B.3 Integrals and series . . . . . . . . . . . . . . . . . . . . . . . . 104

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Script ”Map Projections”

C Formulary

107

C.1 Scale distortion . . . . . . . . . . . . . . . . . . . . . . . . . . 107 C.2 Area distortion . . . . . . . . . . . . . . . . . . . . . . . . . . 109 C.3 Distortion of angles . . . . . . . . . . . . . . . . . . . . . . . . 110 C.3.1 Distortion of the angle element . . . . . . . . . . . . . 111 C.3.2 Distortion of the azimuth D Additional explanations

. . . . . . . . . . . . . . . . 112 114

D.1 The Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 D.2 Alignment of the surfaces . . . . . . . . . . . . . . . . . . . . . 115 D.3 Projection on a coinciding plane . . . . . . . . . . . . . . . . . 116 D.4 Projection on a cone surface . . . . . . . . . . . . . . . . . . . 117 D.5 Projection on a cylinder surface . . . . . . . . . . . . . . . . . 119 D.6 Tissot’s ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 D.7 Specific projections . . . . . . . . . . . . . . . . . . . . . . . . 121

Nomenclature α, δ . . . . . . . . . . azimuthal coordinates β . . . . . . . . . . . . azimuth or grid bearing (datum surface) β  . . . . . . . . . . . . azimuth or grid bearing (image surface) Λ . . . . . . . . . . . . geographic longitude (ellipsoid of revolution) λ, ϕ . . . . . . . . . . geographic longitude/latitude (sphere) ω . . . . . . . . . . . . distortion of the azimuth Ωm . . . . . . . . . . . maximum distortion of an angle ωm . . . . . . . . . . . maximum distortion of the azimuth τ . . . . . . . . . . . . . angle between the parameter curves τ . . . . . . . . . . . . . intersection angle of the parameter curves (datum surface) τ  . . . . . . . . . . . . intersection angle of the parameter curves (image surface) Θ . . . . . . . . . . . . geographic latitude (ellipsoid of revolution) x . . . . . . . . . . . . . position vector(datum surface) x . . . . . . . . . . . . position vector(image surface) a, b . . . . . . . . . . . semi minor and/or semi major axis of the ellipse of distortion C . . . . . . . . . . . . integration constant c . . . . . . . . . . . . . scale factor of the y  -axis (cylinder projection) v

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Script ”Map Projections”

E  , F  , G . . . . . first Gaussian fundamental quantities (projection surface) E, F, G . . . . . . . first Gaussian fundamental quantities (datum surface) H . . . . . . . . . . . . principal point h . . . . . . . . . . . . . distortion along the parameter curve J . . . . . . . . . . . . Jacobian matrix k . . . . . . . . . . . . . distortion along the parameter curve M . . . . . . . . . . . center of the sphere (used in figures) m . . . . . . . . . . . . scale distortion mu , mv . . . . . . . scale distortion along the parametric curves (Gaussian parameters) N . . . . . . . . . . . . north pole (only used in figures) N . . . . . . . . . . . . surface normal (also normal vector; used in formulas) R . . . . . . . . . . . . radius of the sphere (earth) r . . . . . . . . . . . . . radius of an arbitrary circle S . . . . . . . . . . . . south pole x . . . . . . . . . . . . . first component of the position vector (datum surface) x . . . . . . . . . . . . first component of the position vector (image surface) y . . . . . . . . . . . . . second component of the position vector (datum surface) y  . . . . . . . . . . . . second component of the position vector (image surface) dO . . . . . . . . . . . infinitesimal area ds . . . . . . . . . . . . infinitesimal line element O . . . . . . . . . . . . finite area u,v . . . . . . . . . . . Gaussian parameters V . . . . . . . . . . . . area distortion

Chapter 1 Introduction 1.1

What are map projections good for?

1.2

Classification

There are different possibilities to classify the map projections. The criteria may be: extrinsic or intrinsic. The first aspect encompasses the nature, the coincidence and the position of the projection surface. The latter refers to the properties and the generation.

extrinsic

intrinsic

class nature coincidence position properties generation

varieties conical secant transverse eqaual area semi-geometric

plane tangent normal equidistant geometric

cylindrical polysuperficial oblique conformal conventional

Table 1.1: The classification of map projections Actually the classes are not mutually exclusive (see exercises).

1

Chapter 2 Fundamentals 2.1

The Gaussian curvilinear parameters u and v z

u = uc

e3 e1 x

v = vc e2

y

An arbitrary surface defined in IR3 can be described by the position vector x(u, v). Its components are ⎧ ⎨ x = (u, v) y = (u, v) (2.1) x(u, v) = ⎩ z = (u, v). The substitution of u = uc = const., delivers x(uc , v) = x(v). This curve is called a v-line. In the other case x(u, vc ), i. e. v = vc = const., we get u-lines.

Figure 2.1: The position Any arbitrary point of the surface requires two independent vectors defining a new surface vector 1 which is the tangent plane . To ensure this precondition the cross product has to fulfill (2.2) xu × xv = N = 0 , i. e. the surface normal N 2 has to be different from zero. All points of a 1

The index indicates the differentiating parameter. E. g. xu denotes the derivation of vector x with respect to u. 2 Sometimes also called normal vector

2

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

3

surface fulfilling the condition above are called regular . A surface consisting of regular points is called analytical . Two analytical surfaces can be mapped to each other, if there exists a relationship between the the parameters u , v  and u, v. u = u (u, v) und v  = v  (u, v) .

(2.3)

Hence, the Jacobian determinant, derived form the Jacobian matrix J  ∂u ∂u    ∂u ∂v  J =  ∂v (2.4) ∂v  ∂u

∂v

is not allowed be zero and thus the equation ∂u ∂v  ∂u ∂v  − = 0 . ∂u ∂v ∂v ∂u

2.2

(2.5)

The line element As shown in the sketch the differential line element can be expressed by d x = xu du + xv dv

xu du

dx

whereas |d x| = ds .

Thus the square of the line element is xv dv

ds2 = xu xu du2 + 2 xu xv du dv + xv xv dv 2 .     E

F

(2.6)

G

The quantities E = xu xu and F = xu xv as well as G = xv xv are called the first Gaussian fundamental quantities 3 . These quantities describe the metric of a surface. If we think of two different surfaces, one datum surface and one image surface, the above Gaussian quantities of the image surface can be expressed in a similar way as E  = xu xu , 3

F  = xu xv

and G = xv xv .

For convenience further on called Gaussian quantities

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Script ”Map Projections”

Example: Exercise 1 Before starting with the calculations let’s have a look at formula A.1. This is the parametric representation of a sphere, where as the parameter u can be treated as the geographic latitude and the parameter v as the geographic longitude. Thus the datum surface is the sphere. Hence there are other parametric representation of the sphere possible. N P u. Equator

substitution of u =

This is obvious if we analyze the adjacent example. The interpretation of u˙ as an angle between the z-axis and position vector to point P leads us to another parametric representation of the sphere: ⎧ ⎨ x = R sin u˙ cos v x(u, ˙ v) = y = R sin u˙ sin v (2.7) ⎩ z = R cos u˙ This kind of parametrization is equivalent to the latter. The π ˙ and the relationship (B.8) again delivers (A.1). 2 −u

item a): We notice in the problem definition that the two position vectors describe two different surfaces. We already analyzed one surface and know that it is a sphere, which is given as x(u, v). The second position vector is x (u , v  ). Hence referring to (2.3) a projection assumes a position vector like x (u, v). Exercise 1 already encompasses the according relationship by (A.3). Applying this we get ⎧  ⎨ x = R cos v . y  = R sin v x (u, v) = ⎩  z = R tan u Further we have to test if equation (A.3) is fulfilled. This leads us to the below Jacobian determinant:   ∂u R ∂u    ∂u = cos2 u ∂v = 0  = R = 0 . ∂v  cos2 u  ∂v = 0 ∂u ∂v = 1 The determinant is different from zero. Hence if u = 90◦ than the fraction is not defined. If we have a look at the third component of x(u, v) we recognize, that the north pole can not be mapped. Still we get an unique solution for all other points. item b): The question about the geometric interpretation is equivalent to the question about the image surface. As we already know the position vector of the image surface we only need to analyze it. Therefore we simply took the projection in the different planes.

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

5

The projection on the x − y − plane (z  = 0) delivers a circle with radius R and it depends only on the parameter v. The projections on the x − z  − plane and on the y  − z  − plane delivers a straight line which is parallel to z  in both cases. Hence the third component only depends on u ∈ [−∞, +∞]. Summing up all facts shows that the projection surface is a cylinder. I. e. the point of the sphere are projected on a cylinder, coinciding with the sphere at the equator.

+z

+R

+y

+x

Figure 2.2: The image surface Figure 2.2 depicts the resulting image surface in the three planes. The coinciding circle of the cylinder has the same radius as the sphere. item c): In this case we get u = ϕ and v = λ thus x(u, v) = x(ϕ, λ). The three-dimensional coordinates of K are: ⎧ ⎨ x = 4138.425 km y = 581.618 km (2.8) xK (8◦ , 49◦ ) = ⎩ z = 4807.500 km . In the same way one gets the Cartesian ⎧ ⎨ x ◦ ◦ y xM (38 , 56 ) = ⎩ z

coordinates of point M : = 2806.941 km = 2193.022 km = 5280.969 km

(2.9) .

On the image surface the according coordinates for K are given by ⎧  ⎨ x = 6308.008 km  ◦ ◦ 886.533 km y = xK (8 , 49 ) = ⎩  z = 7327.847 km and

item c):

⎧  ⎨ x  ◦ ◦ y xM (38 , 56 ) = ⎩  z

= = =

5019.629 km 3921.764 km 9443.913 km

.

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The shortest distance between the two points K and M on the surface of the sphere 90−φK is the great circle. It can be calculated based on the spherical law of cosine (see B.12):



 K cos s = cos 90◦ − ϕK cos 90◦ − ϕM 



+ sin 90◦ − ϕK sin 90◦ − ϕM cos Δλ

Δλ

90−φ

M

M s

= sin ϕK sin ϕM + cos ϕK cos ϕM cos Δλ s = 2153.521 km

¯ Figure 2.3: The distance KM

The Euclidean distance is deliverd by (2.8) and (2.9) to se ≈ 2143 km. As expected the Euclidean distance is shorter than the great circle. item d): If we develop the surface of the cylinder into the plane, the length of s is given

+y

+x K’

M’

Figure 2.4: The mapped curve in ground- and lateral view by

s =



(R · Δλ)2 + Δz 2 = 3949.952 km

as the scetch depicts. Thus the shortest distance between datum and image surface is distorted by m = s /s = 1.834 4 4

Note: The shortest distance on the image plane is not identical with the mapped image of the great circle. Insofar the comparison between the two curves on the different surfaces is not exact. We discuss this later on.

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item e): The directional derivatives are: ⎧ ⎨ xϕ yϕ xϕ = ⎩ zϕ

= = =

−R sin ϕ cos λ −R sin ϕ sin λ R cos ϕ

(2.10)

⎧ ⎨ xλ y xλ = ⎩ λ zλ

= = =

−R cos ϕ sin λ R cos ϕ cos λ 0 .

(2.11)

and

Being regular (2.2) requires that the normal vector N of a point is not equal to zero. N is ⎧ ⎨ x = −R2 cos2 ϕ cos λ . y = −R2 cos2 ϕ sin λ N = xϕ × xλ = ⎩ 2 z = −R sin ϕ cos ϕ factoring out R2 we get

⎞ cos2 ϕ cos λ N = −R2 ⎝ cos2 ϕ sin λ ⎠ sin ϕ cos ϕ ⎛

.

Thus N (45◦ , 50◦ ) = 0 and P is regular. item f): With (2.10) and (2.11) we already know the directional derivative. Calculating the dot product based on this derivatives we obtain the Gaussian quantities for the sphere: xϕ xϕ = R2 sin2 ϕ cos2 λ + R2 sin2 ϕ sin2 λ + R2 cos2 ϕ or E = R2

as well as

(2.12)

2

2

xϕ xλ = R sin ϕ cos ϕ sin λ cos λ − R sin ϕ cos ϕ sin λ cos λ F

= 0 and 2

2

2

2

(2.13) 2

2

2

2

xλ xλ = R cos ϕ sin λ + R cos ϕ cos λ G = R cos ϕ

or

with

.

(2.14)

Because F = xϕ xλ = |xϕ | |xλ | cos τ = 0   =0

=0

the directional derivatives (which are vectors as well) are intersecting with τ = 90◦ , i. e. the parameter curves are orthogonal.

8

2.3

Script ”Map Projections”

Intersection angles of surface curves As shown by figure (2.5) the line element of two different curves can be depicted as follows

T

β ds

δs

δ

|d x| = ds |δ x| = δs .

Figure 2.5: Intersection of curves

Based on the directional derivative we get d x = xu du + xv dv δ x = xu δu + xv δv

and

and thus the intersection angle γ of the two surface curves is given by ( xu du + xv dv)( xu δu + xv δv) d xδ x = |d x||δ x| ds δs xu xu duδu + xu xv duδv + xv xu dvδu + xv xv dvδv = dsδs Eduδu + F (duδv + dvδu) + Gdvδv . = dsδs

cos γ =

(2.15)

In the same way the azimuth β of an surface curve can be found. As depicted in figure (2.5) the relationship between the azimuth β and the two directional derivatives along the parametric lines (tangent vectors) in point P is given by d x xu cos β = |dx| | xu | and can be simplified by replacing |dx| with cos β =

( xu du + xv dv) xu E du + F dv √ √ = ds E ds E

.

(2.16)

The intersection angle τ between the two parametric lines can be calculated based on the inner product too. One gets (see figure (2.5)) cos τ =

F xu xv =√ √ |xu | |xv | E G

.

(2.17)

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In addition to the above derived equations also the sine and the cosine of the relating intersection angles are shown: √ EG − F 2 (duδv − dvδu) sin γ = (arbitrary surface curves)(2.18) ds δs √ EG − F 2 dv √ (azimuth (bearing)) (2.19) sin β = Eds √ EG − F 2 √ √ sin τ = (parameter curves) (2.20) E G

2.4

The infinitesimal area

The infinitesimal area dO, which is represented by a differential parallelogram, can be computed by the cross product dO = | xu du × xv dv| . Or based on (2.1) the small area could be expressed as dO = | xu du × xv dv| = | xu × xv |du dv = |N| du dv . In combination with Lagrange’s formula we get ( xu × xv )2 = x2u x2v − ( xu xv )2 = EG − F 2 and thus expressed by the Gaussian quantities the infinitesimal parallelogram becomes √ dO = EG − F 2 du dv . The integration of the over the entire surface O delivers   √ EG − F 2 du dv . O= u

v

Expecting the same for the projection surface yields   √  E  G − F 2 du dv . O = u

v

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Script ”Map Projections”

The above result could be applied to an easily understandable example. The Gaussian quantities E, F and G for the sphere (see (2.12) - (2.14)) deliver for the surface   O= R2 cos ϕ dϕ dλ . ϕ

λ

The integration for ϕ ∈ [0, π/2] and λ ∈ [0, π], should yield the area of the hemisphere. Since the order of the integration could be swapped, it is of no importance, which of the parameters is the first. Taking the latitude we get R

2



π/2 0

and next O=R

2

cos ϕ = R2 [sin ϕ]0

 0

π/2

2π

= R2

2 dλ = R2 [λ]2π 0 = 2πR

which is in fact the surface of the hemisphere.

,

Chapter 3 Theory of Distortions 3.1

The scale distortion

The scale distortion

ds ds is the distortion of the line element. It depends on the direction and thus on dv/du. A more detailed description of the scale distortion regarding the results in section (2.2) is given by m=

m2 =

ds2 E  du2 + 2F dudv + G dv 2 = ds2 Edu2 + 2F dudv + Gdv 2

.

(3.1)

Thus the scale distortions along the parameter curves are m2u = h2 =

E E

and m2v = k 2 =

G G

.

The distortions are abbreviated as follows: mu = h (distortion along the u curve) and mv = k (distortion along the v curve). Example: Exercise 1 First the Gaussian quantities of the projection surface have to be calculated 1 . Therefore the tangent vectors along the parameter curves are derived. 1

Note: Instead of the parameters u and v the geographic latitude ϕ and the geographic longitude λ are used

11

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After the derivation we get:  xϕ

=

and

 xλ

=

x y

= R = −R λ sin ϕ

x y

= 0 = R cos ϕ

(3.2)

,

(3.3)

based on this result the Gaussian quantities are E  = xϕ xϕ = R2 + R2 λ2 sin2 ϕ = R2 (1 + λ2 sin2 ϕ) F  = xϕ xλ = −R2 λ sin ϕ cos ϕ and G = xλ xλ = R2 cos2 ϕ . The scale distortion along the parametric curves can be written as √  E R2 (1 + λ2 sin2 ϕ) √ = 1 + λ2 sin2 ϕ mϕ = h = √ = E R2 √ R2 cos2 ϕ G mλ = k = √ = =1 G R2 cos2 ϕ and the scale distortion along the meridian in point P(30◦ north, 45◦ east) is mϕ (30◦ , 45◦ ) = h = 1.07434 . Whereas the circle of latitude is projected in its original length (mλ = 1). Next the scale distortion is examined, whereas mainly the relation with the azimuth is of interest. The scale distortion can be written as m2 =

 2   2  ds2  du  du dv  dv + G = E + 2F ds2 ds ds ds ds

.

If we further regard the formulas found in section (2.3), in addition to (2.19) and (2.16) we get √ sin β =

EG − F 2 dv √ ds E

⇒

√ dv E =√ sin β ds EG − F 2

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13

as well as Edu + F dv √ combined with the relationship above ds E du cos β F sin β = √ −√ √ , ds E E EG − F 2

cos β =

which leads us to E EF  − F E  cos2 β + 2 √ sin β cos β E E EG − F 2  EG − 2F F  + GE  E   − + sin2 β . EG − F 2 E

m2 =

(3.4)

Based on the above formula the following simplifications are introduced E E EF  − F E  √ B = E EG − F 2 EG − 2F F  + GE  E  C = − EG − F 2 E A =

leading to

,

m2 = A cos2 β + 2B sin β cos β + C sin2 β

.

(3.5)

This equation delivers, that the scale distortion depends not only on the Gaussian quantities and thus on the parameters u and v but also on the azimuth denoted as β. m = f (u, v, β) . (3.6) This leads us to the question if there exists a maximal and a minimal distortion respectively. Therefore we set the differential quotient to zero. dm =0 . dβ

(3.7)

Thus we derivate the implicit function of equation (3.5) which yields 2mdm = −2A cos β sin βdβ + 2B cos2 βdβ − 2B sin2 βdβ + 2C sin β cos βdβ and in combination with (B.4) as well as (B.5)) we get 2m

dm = −A sin 2β + 2B cos 2β + C sin 2β dβ

.

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Script ”Map Projections”

Based on this formula we calculate the extreme values. − A sin 2β + 2B cos 2β + C sin 2β = 0

(3.8)

and after some basic derivations (A − C) sin 2β = 2B cos 2β which yields finally for β tan 2β =

2B A−C

.

Thus we got the extreme distortion of the azimuth β. As the tangent is ambiguous, the formula delivers 2βh2 = 2βh1 ± π

,

or

π . 2 This equation reveals that the directions of the maximum/minimum distortion are perpendicular. These two directions are called the principal directions2 and the according distortions are the principal distortions. The re-substitution of A, B and C by the Gaussian quantities delivers after some deductions √ 2(EF  − F E  ) EG − F 2 tan 2βh = (3.9) 2F (EF  − F E  ) + E(GE  − EG ) βh2 = βh1 ±

In exact the same way we get the principal directions of the projection surface. We start with  du 2  du dv   dv 2 1 2 + 2F etc. +G μ = 2 =E m ds ds ds ds In the end we get for the projection surface π βh 2 = βh 1 ± 2 and in detail for tan 2βh tan 2βh 2

√ 2(F E  − EF  ) E  G − F 2 = 2F (F E  − EF  ) + E  (EG − GE  )

Also called base directions

.

(3.10)

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15

The result of formula (3.9) and formula (3.10) comprises, that the principal directions on the datum surface as well as on the projection surface are perpendicular. Hence the principal directions are normally the only mutual perpendicular directions. In our next step we assume F ≡ F  ≡= 0 which means, that the parameter system of the datum surface as well as that of the image surface are perpendicular. In this case, the parameter system is also the system with the principal directions. The above mentioned aspects lead us to the question, which are the absolute values of these distortions. Therefore we modify the two formulas (3.5) and (3.8) which delivers (A cos β + B sin β) cos β + (B cos β + C sin β) sin β = m2 as well as −(A cos β + B sin β) sin β + (B cos β + C sin β) cos β = 0 . If we multiply the first with cos β and the second with sin β and afterwards subtract the second from the first we get (A cos β + B sin β) cos2 β + (A cos β + B sin β) sin2 β = m2 cos β or

A cos β + B sin β = m2 cos β

.

(3.11)

In our next step we multiply equation (3.5) and (3.8) with sin β and cos β respectively. The result is (A cos β + B sin β) sin β cos β + (B cos β + C sin β) sin2 β = m2 sin β −(A cos β + B sin β) sin β cos β + (B cos β + C sin β) cos2 β = 0 . Then we add the equations and obtain B cos β + C sin β = m2 sin β

.

(3.12)

We transpose the formulas (3.11) and (3.12) and get (m2 − A) cos β = B sin β and (m2 − C) sin β = B cos β . The multiplication of the equations delivers a quadratic expression of m. This term does not depend on β. Is is (m2 − A)(m2 − C) = B 2

16

Script ”Map Projections”

or

m4 − (A + C)m2 + AC − B 2 = 0 ,

from which we get the two solutions  A + C A + C 2 + − AC + B 2 m21 = 2 2  A+C A + C 2 2 − m2 = − AC + B 2 2 2

und

(3.13)

.

(3.14)

The first solution is the square of the first principal direction and is abbreviated by a2 , whereas the square of the second principal direction is denoted by b2 . This means that m21 = a2

und m22 = b2

.

From this we derive further quantities. The first is called medial distortion U 2 and is defined as follows 2U 2 = a2 + b2 = A + C

.

The second is the area distortion V , which is defined as  A + C 2  A + C 2 V 2 = a2 b2 = − + AC − B 2 = AC − B 2 2 2

(3.15)

.

(3.16)

Re-substituting the quantities A, B and C by the Gaussian quantities delivers 2U 2 = a2 + b2 = A + C = =

EG − 2F F  + GE  EG − F 2

E  EG − 2F F  + GE  E  + − E EG − F 2 E

and V 2 = a2 b2 = AC − B 2 E   EG − 2F F  + GE  E    EF  − F E  2 √ − = − E EG − F 2 E E EG − F 2

,

for the area distortion respectively, which can be simplified to V2 =

E  G − F 2 EG − F 2

.

(3.17)

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17

In the special case of F ≡ 0, e. g. the parameter system of the datum surface is orthogonal, both quantities can be written as E  G EG + GE  = + = h2 + k 2 and EG E G F 2 E  G F 2 − = h2 k 2 − = a2 b2 . = EG EG EG

2U 2 = V2

(3.18) (3.19)

If further the parameter system of the projection surface is also orthogonal (F  ≡ 0) we get (3.20) V 2 = h2 k 2 = a2 b2 as well as

E G and b2 = k 2 = E G Formula (3.4) can then be simplified to a2 = h2 =

m2 = a2 cos2 β + b2 sin2 β

3.2

.

.

(3.21)

(3.22)

The distortion of areas

According to the definition of the scale distortion, the area distortion V is given by d O . (3.23) V = dO E. g. the area distortion is the ratio of the infinitesimal area of the datum surface and the infinitesimal area of the image surface. Both quantities are already introduced in section 2.4. Applying to the terms of this section we get E  G − F 2 V2 = (3.24) EG − F 2 combined wiht (3.17) delivers V = a b . Thus the area distortion can be derived by simply multiplying the principal distortions. This was already shown in the section above. But there is still another method to determine the distortion of areas based on the tangent vectors and the enclosed angle. It is √ √ √ √ d O = E G sin τ du dv and d O  = E  G sin τ  du dv

18

Script ”Map Projections”

considering that the distortion along the parameter curves is abbreviated by h and k we obtain sin τ  V = hk . sin τ If the parameter system of the datum surface as well as of the projection surface is orthogonal the angle of intersection of the parameter lines is 90◦ and thus τ = τ  = π/2 which means that the principal directions coincide with the parameter lines. By applying (3.24) we get V = hk = ab .

3.3

The distortion of the azimuth

The distortion of the azimuth ω is the difference between the azimuth β  on the image surface and the according azimuth β on the datum plane. Hence we get ω = β − β . During the next steps the relationship between the azimuth and the distortion of the azimuth are examined. We assume further F ≡ F  ≡ 0. From equation (2.16) and (2.19) we get, considering the precondition F  = 0 cos β  =

√

E

du ds ds ds

and

sin β  =

√

G

dv ds , ds ds

regarding m=

ds ds

as well as

du cos β = √ ds E

and

dv sin β = √ ds G

we derive b sin β and m a cos β  = cos β . m

sin β  =

(3.25) (3.26)

The division of the above formulas leads us to a rather simple equation for the relation between the azimuth of the datum and the image surface tan β  =

b tan β . a

(3.27)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch Equation

sin ω = sin(β  − β) and

19

cos ω = cos(β  − β)

in combination with the addition theorem for the sine and the cosine (see (B.1) as well as (B.2)) leads us to sin ω = sin β  cos β − cos β  sin β cos ω = cos β  cos β + sin β  sin β

and .

The substitution of cos β  and sin β  with the term (3.25) and (3.26) delivers an equation for the distortion of the azimuth the two formulas b a b−a sin β cos β − sin β cos β = sin β cos β and m m m b 1 a cos2 β + sin2 β = (a cos2 β + b sin2 β) . cos ω = m m m sin ω =

For the tangent of the azimuth the same deductions can be performed. We get tan(β  − β) = tan ω in combination with (B.3) tan ω =

tan β  − tan β 1 + tan β  tan β

and by applying (3.27) the term can be simplified as follows tan ω =

3.4

b a

tan β − tan β (b − a) tan β . = b 2 a + b tan2 β 1 + a tan β

(3.28)

The infinitesimal angular distortion

The infinitesimal distortion can be expressed as follows dβ  =? dβ Equation (3.27) already denotes the relation between β  and β. Thus it is sufficient to derivate the formula. Doing this we get b 1 1  dβ dβ = cos2 β  a cos2 β

20

Script ”Map Projections”

in combination with (3.26) we find a2 m2

b 1 1 dβ  = dβ 2 a cos2 β cos β

,

by reducing the fraction by cos2 β and further simplifications we can derive dβ  2 m = ab . dβ

(3.29)

Thus we found a relation between the distortion of the infinitesimal angular distortion, the scale distortion and the principal distortions. Based on the above formula we may be interested in the question: In which direction the infinitesimal angle is not distorted. Thus formula dβ  =1 dβ has to be fulfilled. With (3.22) we get for (3.29) the term a2 cos2 βm + b2 sin2 βm = a b , which yields by substitution of the sine and the cosine and factoring out cos2 βm (a2 − b2 ) = a b − b2 or cos2 βm =

a b − b2 b ⇒ = a2 − b2 a+b

and thus

tan2 βm =

Based on tan β  =

a b

sin2 βm =

a a+b

.

b tan β a

one gets  = tan2 βm

b a

⇒

 tan βm tan βm =1

e. g.  = βm + βm

π 2

.

(3.30)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

3.5

21

The maximal distortion of the azimuth

Based on the definition for the distortion of the azimuth ω = β  − β we try to find out, if there are directions where the distortion is maximal. By differentiation we get dβ  dω = −1 =0 . dβ dβ 

Obviously this is the direction with dβ = 1, which is βm , e. g. the direction dβ  − βm where the infinitesimal angle is not distorded. Based on ωm = βm together with formula (3.28) the amount of the maximum distortion of the azimuth can be expessed by tan ωm =

(b − a) tan βm a + b tan2 βm

.

By applying the specific equation for the direction with the maximum distortion for the infinitesimal angle (see 3.30) we get (b − a) ab b−a tan ωm = . (3.31) = √ 2a 2 ab It is possible to deduce similar equations for the sine and the cosine of ωm These are √ b−a 2 ab and cos ωm = . sin ωm = b+a b+a The maximal distorted angle Ωm is due to symmetry Ωm = 2 |ωm| , because the maximal distorted azimuth can be found in the first and in the fourth quadrant.

3.6

Ellipse of distortion

Basic work in the field of map projections was done by Nicolas Auguste Tissot (1824 - 1890). This is the reason why in many textbooks and other related literature the ellipse of distortion is also calledTissot’s Indikatrix.

22

Script ”Map Projections”

Starting point for the next examinations is the question how the infinitesimal circle with the radius ds is mapped on the image surface. Further F ≡ F  ≡ 0 is assumed The infinitesimal line element can be described on both the datum and the image surface as ds2 = Edu2 + Gdv 2 = dξ 2 + dη 2 and ds2 = E  du2 + G dv 2 = dξ 2 + dη 2 , whereas for the datum surface √ ds cos β dξ = ds cos β = E du ⇒ du = √ E √ ds sin β dη = ds sin β = G dv ⇒ dv = √ G

as well as

and for the image surface by applying (3.34) and (3.35) √ dξ  = ds cos β  = E  du = ds a cos β √ dη  = ds sin β  = G dv = ds b sin β

(3.32) (3.33)

(3.34) (3.35)

(3.36) (3.37)

After the division by ds, which delivers the radius of the unit circle we get for the image plane dξ  = m cos β  = a cos β ds dη  = m sin β  = b sin β = ds

ξ = η and thus

as well as

(3.38) (3.39)

ξ 2 + η 2 = m2

in combination with

ξ 2 η 2 + 2 =1 . a2 b This is the equation of the so called distortion ellipse or Tissot’s Indikatrix. The above formula delivers further, that the infinitesimal unit circle is mapped in its affine image e. g. the ellipse. At this point we can draw some conclusions. The three most important are stated below.

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23

1. The scale distortion m is the length of the radius vector of the distortion ellipse. For two arbitrary direction which are perpendicular β and β = β + π/2 the scale distortion m is 

ξ  = a cos β ; ξ = a cos(β + π/2) = −a sin β η  = b sin β ; η  = b sin(β + π/2) = b cos β

(3.40) (3.41) (3.42)

squared and added delivers ξ 2 + η 2 = m2 = a2 cos2 β + b2 sin2 β 2

2

ξ +η = m and thus

2

2

2

2

2

= a sin β + b cos β

m2 + m2 = a2 + b2

(3.43) (3.44)

.

2. For the distortion of areas the distortion ellipse delivers a geometrically based interpretation which is as follows V =

d O Area of the Indikatrix abπ = = = ab . dO Area of the unit circle π

3. As far as the distortion of the angles is concerned Tissot’s Indikatrix depicts (the principal distortions: m1 ¡ m2 ) for a ray rotating clockwise d β >1 dβ starting at ξ = ξ  = 0 until d β =1 dβ is reached and thus the maximum distortion of the azimuth ωm . Subsequently we get d β 0 or (a2 − 1)(1 − b2 ) > 0 are fulfilled. This is the case if either 1. a2 > 1 und b2 < 1 or 2. a2 < 1 und b2 > 1. If a2 = 1 or b2 = 1 the two directions diminuish to one. Denoting the equidistant curves with βl and βl and assuming F ≡ F  ≡ 0, we get  dv 2 E a2 − 1 E − E = = m2 = −  . du G −G G 1 − b2 Together with the two equations cos β du = √ ds E one obtains

sin2 βl G cos2 βl E

=

and

dv sin β = √ ds G

E a2 − 1 a2 − 1 2 = tan β = l G 1 − b2 1 − b2

and for the image plane tan2 βl = Example (Exercise 9 d)):

b2 b2 a2 − 1 2 tan β = l a2 a2 1 − b2

.

26

Script ”Map Projections” H

u’

P

R 90−u M

P’

u

The term equidistant, which is in use for a certain class of projections, characterizes the property, that an array of curves crossing the central point or its image respectively are mapped equidistant.

Based on the general formula for azimuthal projections , which can be characterized by   x = u cos v     x (u , v ) = y  = u sin v  Figure 3.1: The equidistant azimuthal projection it is necessary to find the relation between u’,u as well as v’ and v. From sketch (see fig. 3.1) we can deduce π  u = R −u and v  = v . 2 Thus we obtain the coordinates of the image by   x = R( π2 − u) cos v  x (u, v) = . y  = R( π2 − u) sin v The derivative with respect to u and v delivers   xu = −R cos v  xu = yu = −R sin v or   xv = −R( π2 − u) sin v  xv = yv = R( π2 − u) cos v

.

The Gaussian quantities are therefore: E  = xu · xu = R2 und F  = xu · xv = 0 and G = xv · xv = R2 ( π2 − u)2. In combination with the quantities of the sphere the principal distortions a and b are     π R2 ( π2 − u)2 −u E R2 G 2 = = = . a= = 1 und b = E R2 G R2 cos2 u cos u Assuming, that the principal point of the projection coincides with the north pole, the parameters u and v can be treated as the geographic latitude and the geographic longitude. The principal distortion for u = 60◦ is given by 1 a = 1 und b = π = 1.0472 . 3 As expected the meridians passing the principal point are all equidistant.

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

3.7.2

27

Equal area projections

A projection can be called equal area, if d O = d O , is fulfilled, e. g. the infinitesimal area of the datum and the image surface are equal. Further we can deduce d O = 1 = ab = V dO

.

(3.46)

Combined with (3.19) we get V 2 = 1 ⇒ E  G − F 2 = E G − F 2 and O=

  √ u

EG −

F 2 du dv



and O =

v

  √ u

E  G − F 2 du dv ,

v

which depicts the property equal area is not only fulfilled in an infinitesimal environment but also in a finite. Because of a b = 1 one of the principal distortions is always less than 1 whereas the other is greater than 1. As a consequence dealing with equal area projections there always exists a direction with a scale distortion of 1. This direction is given by (datum surface) tan2 βl =

a2 − 1 a2 − 1 = = a2 1 − b2 1 − a12

or tan βl = ±a and tan2 βl =

b2 a2 − 1 b2 1 − b2 = = b2 ⇒ tan βl = ±b 2 1 2 2 b a 1−b b2 1−b2

which is the according direction to the image plane. Due to tan2 βm =

a b

und

 tan2 βm =

b a

 1 1 a = bzw. b = b a

28

Script ”Map Projections”

these directions βl and βl indicate also the maximal distortion of the azimuth. Thus it is tan βm = tan βl = ±a and

 tan βm = tan βl = ±b .

The absolute value of the maximal distorted azimuth ωm is given by  π ω  a m tan − = = a and (3.47) 4 2 b  π ω  b m + = =b. (3.48) tan 4 2 a

3.7.3

Conformal projections

According to Carl Friedrich Gauß a projection can be called conformal if the image is similar to the original figure. This means that e. g. the unit circle on the datum surface delivers a circle on the image surface. In general the latter is actually no longer a unit circle. Thus the scale distortion is a=b=m or

(3.49)

G E = (assuming F ≡ F  ≡ 0). . E G It is an intrinsic property of conformal projections, that the scale distortion depends not on the azimuth.

Chapter 4 Oblique map projections 4.1

From Gaussian parameters to geographic coordinates

Till this section the parameters u and v could be treated as ϕ and λ which denote the geographical latitude and longitude respectively as long as we stick to the normal alignment of the two surfaces. Thus u = ϕ and v = λ and furthermore we can derive the position vector x = (u, v) and x = (ϕ, λ) in a rather easy way. Hence things change if the north pole remains no longer the central point of the projection (see fig. below). H

P’

N 90 − φH

P 90−u u

Δλ

N

H

M

90 − φ P P

v 90 − u

(a) Central point and north pole

(b) The polar triangle

Figure 4.1: The oblique alignment of the two surfaces Normally the position on the sphere of an arbitrary point P are given in geo29

30

Script ”Map Projections”

graphic coordinates ϕ and λ, whereas the projection formulas use the Gaussian parameters u and v. This makes sense because as already mentioned the latter formulas depend not on the attitude of the central point however they refer to it. This means that we have to find a relationship between the Gaussian parameters and the geographic coordinates. This relationship is not linear as figure 4.1 depicts. As figure 4.1 depicts we need a polar triangle to find the above mentioned relationship. To explain this in detail, we take a azimuthal projection as an example. It is further assumed, that the central point is located at H(ϕH , λH ). Furthermore we take an arbitrary point P (ϕP , λP ). Our next step leads us to the spherical law of cosines cos(90◦ −u) = cos(90◦ −ϕH ) cos(90◦ −ϕP )+sin(90◦ −ϕH ) sin(90◦ −ϕP ) cos(Δλ) which delivers the relationship between u and the geographical coordinates. The above equation can be simplified to sin(u) = sin(ϕH ) sin(ϕP ) + cos(ϕH ) cos(ϕP ) cos(Δλ) . The second parameter v can be determined by the spherical law of sines – with respect to the central point H – as shown by sin(Δλ) sin(v) = bzw. cos(ϕP ) cos(u) sin(Δλ) cos(ϕP ) . sin(v) = cos(u)

4.2

The azimuthal coordinates

Dealing with oblique projections it makes sense to change the parameter system once again. Although this step isn’t carried out by all authors it makes map projections much easier as shown below. Looking at the spherical triangle with its edges in N, H and P and the appropriate great circles between the points we introduce a local azimuth α (counted clockwise) which is defined by the meridian through H and the great circle through H and P . The second parameter is the polar distance δ. This parameter is defined as the distance of the great circle (or the orthodrome)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

31

counting from the central point H to P . Henceforth the new parameters α and δ replace the Gaussian parameters u and v. The central point H – with arbitrary attitude – to which both parameters refer is like a pole to the new coordinate system. Thus the circles δ = const. and the lines α = const. can be seen as a generalization of the parallel circles and the meridians respectively. The new parameters deliver considering the position vector ⎧ ⎨ x = y = x(δ, α) = ⎩ z =

the sphere the below equations for R sin δ cos α R sin δ sin α R cos δ.

(4.1)

The related Gaussian quantities are E = R2

sowie F = 0 und G = R2 sin2 δ

.

As already outlined the central point is like a pole for the newly introduce parameter system. Thus the definition of this point is crucial. Naturally the central point depends on the projection surface which can be a cone a cylinder or a plane. The latter is the easiest because the central point is the coincidence of the plane and the sphere (if the plane is tangent). In the case of the cylinder surface the central point is defined as the intersection of the axis of the cylinder and the surface of the sphere. In almost the same way we get the central point considering a conical projection. In this context it is the intersection of the line between the apex of the cone and the center of the sphere.

4.3 4.3.1

Specific curves on the datum surface The loxodrome

The curve intersecting each meridian between two points P1 and P2 at a constant azimuth is called loxodrome1 . This curve was of great importance in ancient navigation. Based on conformal maps the course angle could be determined by geometric means, because the loxodrome is the shortest 1

Sometimes also called rhumb line

32

Script ”Map Projections”

distance between two points in such a map. Thus the course angle e. g. for navigation on a ship could be easily derived by a rule. cBased on figure 4.2 we can conclude dλ dϕ

or

tan β dϕ cos ϕ and by integration between P1 (ϕ1 , λ1 ) and P2 (ϕ2 , λ2 ) and solving for tan β

ds

R db

tan β = cos ϕ

dλ =

β R cos b dl

Figure 4.2: The differential triangle



tan β = ln tan

π 4

λ2 − λ1   + ϕ22 − ln tan π4 +

ϕ1 2



(4.2) we get the course angle between the two points. The distance along the loxodrome can be deduced (see fig. 4.2) from cos β = 

or

ϕ2

s= ϕ1

Rdϕ ds

and thus ds =

Rdϕ cos β

R R dϕ = (ϕ2 − ϕ1 ) . cos β cos β

(4.3)

Exercise 11 Based on a map which is the result of a conformal projection of the ellipsoid of revolution to the plane a captain derives the course angle of his passage from San Juan (Puerto Rico) (ϕJ = 18.28◦ , λJ = 66.07◦) to Lisbon (ϕL = 38.43◦ , λL = 9.08◦ ). The course angle shall be the same for the whole journey. a) Calculate the difference of the course angle, if the map is not derived from a ellipsoid of revolution but from a sphere (radius = 6 370 km) b) How is the difference of the length of the loxodrome between the ellipsoid of revolution and the sphere? What is the length of the loxodrome if we calculate with the course angle of the sphere? c) And finally, what is the difference in the distance between the loxodrome and the shortest distance on the sphere?

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33

General information: a = 6377.4 km; e2 = 0.0066744        e   x π π Φ 1 − e sin Φ 2 dx M(Φ) = ln tan + , dΦ = ln tan + · cos x 2 4 N(Φ) cos Φ 4 2 1 + e sin Φ Item a) Referring to (4.2) one gets the course angle from San Juan to Lisbon on the sphere βK = 67◦ 9471. The loxodrome on the surface of the ellipsoid of revolution: In combination with the Gaussian quantities (E = M 2 (Φ); F = 0; G = N 2 (Φ) cos2 (Φ)) we get for the course angle on the ellipsoid of revolution in analogy to figure 4.2 tan βE =

N(Φ) cos(Φ)dΛ M(Φ)dΦ

.

This ordinary differential equation can be solved by seperation of the variables M(Φ) dΦ = dΛ tan βE N(Φ) cos(Φ) and integrated afterwards (limits J: San Juan L: Lisbon)  L  L M(Φ) dΦ = tan βE dΛ J N(Φ) cos(Φ) J or ΛL −ΛJ = tan βE



π

 π Φ  1 − e sin Φ  2e  ΦL  1 − e sin ΦL  e2 J J ln tan + + −ln tan 4 2 1 + e sin ΦL 4 2 1 + e sin ΦJ

solved to tan βE delivers the course angle on the ellipsoid of revolution βE 

tan βE = ln tan

π 4

+

ΦL 2



ΛL − ΛJ  e 1−e sin ΦL 2 π − ln tan + 1+e sin ΦL 4

ΦJ 2



1−e sin ΦJ 1+e sin ΦJ

 e2

. (4.4)

If we apply the appropriate figures we get βE = 68◦0490 .

34

Script ”Map Projections”

Thus the difference between the two solutions equals Δβ = βE − βK = 0◦ 1019 . Item b) The length of the loxodrome on the surface of the sphere can be solved immediately by formula (4.3). The equation yields sK = 5 992.894 km . The differential triangle on the ellipsoid of revolution delivers the course angle or azimuth respectively based on the Gaussian quantities cos β =

M(Φ)dΦ ds

.

Paralleled to the sphere, we get an ordinary differential equation which leads after the separation of the variables to ds =

M(Φ)dΦ cos β

which can be integrated in the next step. One gets  ΦL  1 a(1 − e2 ) ΦL 1 M(Φ) dΦ = s=  3 dΦ . (4.5) cos β ΦJ cos β 2 ΦJ 2 1 − e sin Φ This equation comprises an elliptic integral of the first kind and cannot be expressed in terms of elementary functions. Thus we need another approach. The first is the solution by numerical integration. One of this approaches is Romberg’s method. Doing so we get the results depicted in table 4.1.2 . Below a rough description of the algorithm: • General principle: Subdivide the integral into trapezoids • Start with one trapezoid • Calculate the area (bounds: [α, β]) (α − β) ∗ (f (α) + f (β))/2 2

Note: These results are only a first approximation!

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

35

Figure 4.3: The function to be integrated • Halve the interval(s) and calculate another/other support point(s) in the middle of each sub-area • As a result the areas are doubled • Calculate the sum and put it in a table (see below) • Derive from the first column a linear combination as follows: Lki (f ) =

1 k−1 (22k Li+1 (f ) − Lik−1 (f )) 22k − 1

• Stop the iteration if the difference of the last two linear combinations is less than a threshold ε otherwise continue with the third step L1i =

4L0i+1 −L0i 3

L2i =

16L1i+1 −L1i 15

L00 =5974.3145 L01 =5973.8062 L10 =5973.6368 L02 =5973.6802 L11 =5973.6381 L21 =5973.6382 L03 =5973.6487 L12 =5973.6382 L22 =5973.6382

L3i =

64L1i+1 −L1i 63

L32 =5973.6382

Table 4.1: Romberg’s-Method applied to 4.5

36

Script ”Map Projections”

The result for the length of the loxodrome sE from point J to point L computed up to the fourth decimal place is sE = 5 973.6382 km . There are still other approaches to solve 4.5. One of these is the series expansion. The mathematical textbooks dealing with integration and series 3 expansion delivers for (1 − x)− 2 the expression 3

(1 − x)− 2 = 1 +

3 15 2 x+ x . . . (|x| < 1) . 2 8

In the above expansion the series was aborted after the third term. We will see that three terms are enough to get a good approximation for the solution of 4.5. The substitution of

x = e2 sin2 Φ

yields for 4.5  a(1 − e2 ) ΦL 3 15 s= (1 + e2 sin2 Φ + e4 sin4 Φ + . . . ) dΦ or cos β 2 8 ΦJ     ΦL a(1 − e2 ) 3 2 ΦL 2 15 4 ΦL 4 s= dΦ+ e sin Φ dΦ+ e sin Φ dΦ +. . . . cos β 2 8 ΦJ ΦJ ΦJ Considering (B.19) and (B.21) we thus are able compute the length sE of the loxodrome on the surface of the ellipsoid of revolution  ! a(1 − e2 ) 3 2 1 1 (ΦL − ΦJ ) − (sin 2ΦL − sin 2ΦJ ) sE = · (ΦL − ΦJ ) + e cos βE 2 2 4 15 4 3 1 e (ΦL − ΦJ ) − (sin 2ΦL − sin 2ΦJ ) 8 8 4 !  1 (sin 4ΦL − sin 4ΦJ ) + . . . . + 32

+

If we insert the appropriate quantities the length of the loxodrome is sE = 5 973.6382 km .

(4.6)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

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Compared with the reference the difference is less than some centimeters. Furthermore the result reveals that the length of the loxodrome on the sphere is almost the same as the length on the ellipsoid. If we compute the length of the loxodrome applying the course angle of the sphere, we get sE = 5 947.396 km .

38

Script ”Map Projections”

4.3.2

The orthodrome The orthodrome is the shortest connection between two points P1 (ϕ1 , λ1 ) and P2 (ϕ2 , λ2 ) on the surface of the sphere. Thus the orthodrome is is identical with the great circle and this is why the length can be calculated with the spherical cosine theorem.

φ2

90° −

λ2 − λ1

− 90°

φ1

N

Pl A 1

P2

With the formula below we get the distance s on the unit sphere

s

cos s = cos(90◦ − ϕ1 ) cos(90◦ − ϕ2 ) + + sin(90◦ − ϕ1 ) sin(90◦ − ϕ2 ) cos(λ2 − λ1 ) Figure 4.4: The orthodrome with the relationship cos(90◦ − x) = sin x the equation can be simplified to cos s = sin ϕ1 sin ϕ2 + cos ϕ1 cos ϕ2 cos(λ2 − λ1 ) . With so = R · s the length so of the orthodrome on the sphere with radius R is derived. The azimuth A1 from point P1 to P2 can also be deduced based from the spherical cosine theorem cos A1 =

sin ϕ2 − sin ϕ1 cos s cos ϕ1 sin s

.

As an alternative this could also be done with the sine theorem as follows sin(λ2 − λ1 ) sin(λ2 − λ1 ) sin A1 = or sin A1 = cos ϕ2 . ◦ sin(90 − ϕ2 ) sin s sin s Hence there is a problem with this approach because the sin is not unique in the interval [0, π]. Thus the mutual location of P1 and P2 has to be considered. For the projection of the orthodrome O further support points are needed. Thus we need another theorem of the spherical trigonometry the cotangent theorem. Considering the polar triangle NP1 P the cotangent theorem yields tan b =

sin(λ − λ1 ) cot A1 + cos(λ − λ1 ) sin ϕ1 cos ϕ1

(P (ϕ, λ) ∈ O).

(4.7)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

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The unknown azimuth A1 can be replaced by applying the theorem once again cos ϕ1 tan ϕ2 − sin ϕ1 cos(λ2 − λ1 ) cot A1 = . (4.8) sin(λ2 − λ1 ) Putting (4.8) in (4.7) delivers after some basic transformations tan b =

tan ϕ1 (cos(λ − λ1 ) sin(λ2 − λ1 ) − cos(λ2 − λ1 ) sin(λ − λ1 )) + tan ϕ2 sin(l − λ1 ) sin(λ2 − λ1 )

Applying the addition theorem for the sine and the cosine yields sin(λ2 − λ) = cos(λ − λ1 ) sin(λ2 − λ1 ) − cos(λ2 − λ1 ) sin(λ − λ1 ) which finally leads us to tan b =

tan ϕ1 sin(λ2 − λ) + tan ϕ2 sin(λ − λ1 ) sin(λ2 − λ1 )

.

After all we found a functional relationship between the geographic longitude λ and the geographic latitude ϕ. We now can take any λ with λ ∈ [λ1 , λ2 ] to determine intermediate points and thus can map the result on the plane. Exercise 11: Item c): The length s of the orthodrome on the surface of the unit circle between San Juan and Lisbon equals cos s = cos(90◦ −ϕJ ) cos(90◦ −ϕL )+sin(90◦ −ϕJ ) sin(90◦ −ϕL ) cos(λL −λJ ) . Thus the length of the orthodrome so on the surface of the sphere with the radius R = 6371 km equals so = R arccos s = 5 905.364 km . Finally the orthodrome is 88 km longer than the loxodrome connecting the two points.

Chapter 5 Map projections with an orthogonal parameter system In this chapter only projections with orthogonal parameter systems are considered. That is to say that both systems – the system of the datum and that of the image – have to be orthogonal. As a consequence the Gaussian quantities F and F  need to be zero. We first start with the analysis of the cone as projection surface.

5.1

Conic projections

Based on geometric considerations we get for the special case for a cone with one standard parallel the below relationships s = rg ε = α sin θ rg and thus ε = α sin θ = α sin

π

,

 − δg = α cos δg

2 which reveals the functional relationship between α and ε.

,

Expressed in polar coordinates the above formula delivers (see appendix D.4 and D.5) r = r(δ) and ε = n α with n = cos δg = sin θ . 40

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

41

Converted to rectangular Cartesian coordinates we get   x = rg − r cos ε  x (δ, α) = y  = r sin ε.

(5.1)

with rg = R tan δg = R cot θ. The derivation of (5.1) with respect to the two parameters yields the Gaussian quantities for the image surface E =

 d r 2 dδ

,

F  = 0 and G = n2 r 2

.

Combined with the quantities of the sphere (E = R2 , F = 0, G = R2 sin2 δ) the principal distortions can be expressed as   E G 1 dr nr a= = and b = = . (5.2) E R dδ G R sin δ As the principal distortions depend only on δ, the distortions on the parallel circles are constant.

5.1.1

Equal area conic projections

Based on the equations expressing the relationships of principal distortions (see (5.2)) the distortion of area can be derived as follows V = ab = 1 =

1 dr nr R d δ R sin δ

.

The separation of the variables delivers r dr = which after integration leads to   R2 r dr = sin δ dδ n

R2 sin δ dδ n

or r 2 = −

2R2 cos δ + C n

.

(5.3)

The above formula reveals the radius as a function of δ or more rigorous r = r(δ). Thus we found a relationship between the parameter system of

42

Script ”Map Projections”

the sphere and the cone. But there are still two more variables or degrees of freedom to define the properties of the projection. Below some examples. Example 1: Imposing one (coinciding) standard parallel yields n = cos δg and further b(δg ) = 1 =

cos δg r(δg ) R sin δg

.

At this point it is advantageous to square the above equation and afterwards inserting the result in (5.3). One gets cos2 δg (−2R2 + C) =1 , R2 sin2 δg resolved to C yields R2 sin2 δg + 2R2 cos2 δg C= cos2 δg or C = R2

2 1 + cos2 δg 2 1+n = R cos2 δg n2

.

Thus finally we found r2 = −

2R2 1 + n2 cos δ + R2 n n2

or

R√ 1 + n2 − 2n cos δ (5.4) n = 0 respectively. This formula shows , that with respect to r(δ = 0) = R 1−n n the principal point is not mapped as a single point. r=

By inserting (5.4) in (5.2) the principal directions which are the main axis of the distortion ellipse are √ nr sin δ 1 + n2 − 2n cos δ b= = and (because of a = 1/b) a = √ R sin δ sin δ 1 + n2 − 2n cos δ

.

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

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Example 2: If we want to map the central point as a single point we impose (based on formula (5.3)) r(δ = 0) = 0. Thus one gets C=

2R2 n

for the integration constant. Introducing C in (5.3) delivers r2 =

2R2 (1 − cos δ) . n

(5.5)

The relationship 1 − cos δ = 2 sin2 ( 2δ ) (see (B.9)) yields together with (5.5) 2R δ r = √ sin n 2

.

And the related principal distortions are (with respect to (B.4))

 √ cos 2δ n

δ . and b = a= √ n cos 2 This case actually delivers no coinciding standard parallel because of cos δg   = 1 . b(δg ) = cos δ2g Hence the variable n can be chosen in such a way that another parallel circle δ = δl is coinciding. Therefore we need δ    √ l 2 δl n = cos or n = cos . 2 2 Example 3 (Albers equal area conic projection): There is a third way to determine the two constants in (5.3). If we define two coinciding parallels we get two equations with conditions for the both variables n and C. Thus we get b(δl1 ) = 1 =

n r(δl1 ) R sin δl1

and b(δl2 ) = 1 =

n r(δl2 ) R sin δl2

44

Script ”Map Projections”

which leads to r(δl1 ) =

R sin δl1 n

as well as r(δl2 ) =

R sin δl2 n

.

Inserting 5.4 into the above equation and squaring the result yields 2 R2 R2 sin2 δl1 cos δl1 + C = n n2 2 R2 R2 sin2 δl2 cos δl2 + C = − n n2

−

and

(5.6)

.

(5.7)

Subtraction (5.7) from (5.6) delivers 2 R2 R2 sin2 δl1 R2 sin2 δl2 2 R2 − cos δl2 − cos δl1 = n n n2 n2 and after some simplifications 2 (cos δl2 − cos δl1 ) =

1 (sin2 δl1 − sin2 δl2 ) . n

Resolved to n one obtains

" # 1 sin2 δl1 − sin2 δl2 n= 2 cos δl2 − cos δl1

and with sin2 δ = 1 − cos2 δ in combination with the third binomial theorem the equation can be modified to 1 n = (cos δl2 + cos δl1 ) . 2

(5.8)

Hence, the integration constant C has to be resolved. Therefore we insert n in equation (5.6) or (5.7). If we take (5.6) we get after some deductions C=

R2 sin2 δl1 2 R2 cos δl1 + n2 n

and further obtain based on a common denominator " # 2 δ 2n cos δ sin R2 l1 l1 + (sin2 δl1 + 2n cos δl1 ) . = C = R2 n2 n2 n2

(5.9)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

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Inserting (5.9) in (5.3) we finally get R2 (sin2 δl1 + 2n cos δl1 − 2n cos δ) or explicitly n2 R sin2 δl1 + 2n(cos δl1 − cos δ) r = n

r2 =

(5.10)

whereas the varialbe n refers to (5.8).

5.1.2

Conformal conic projections

The scale distortion of conformal projections does not depend on the azimuth but only on the location and thus on the Gaussian parameters. Expressed by azimuthal coordinates α and δ the scale distortion m is m = f (δ, α) . Hence, the infinitesimal unit circle is mapped as a circle again, but as as a circle with a different radius of course. To get a conformal projection m=a=b has to be fulfilled. Taking the principal distortion of equation (5.2) one gets after some deductions dδ 1 dr = n . r sin δ In this equation the variables are already separated and thus could be integrated. The integration delivers ln r = n ln tan

δ + ln C 2

or ln r = ln C tann which leads us to

,

δ 2

δ . (5.11) 2 This formula encompasses the relationship of the radius and the according parameter δ and thus r = r(δ) is solved. From this formula we can conclude r = C tann

46

Script ”Map Projections”

that the principal point is in fact a point, because r(δ = 0) = 0. Furthermore the integration constant is like a scale factor. In addition this formula comprises the two parameters n and C based on which we are able to introduce further constraints. This is explained by the following examples. Example 1: The first possibility to introduce a constraint with a geometrical interpretation is to postulate that the coinciding parallel should not be distorted. The mathematical constraint is b(δg ) = 1 . Applying this in (5.2) than we obtain together with (5.11) C=

R sin δg n tann

δg 2

.

By geometrical interpretation we see n = cos δg and thus the relationship is simplified to R tan δg . C= tann δ2g Finally the radius r of a conformal conic projection with one coinciding standard parallel is R tan δg n δ r= . tan δ g 2 tann 2 Example 2: The alternative is to impose two coinciding standard parallels. If we do so, we get for C the constraint (in combination with b(δ1 ) = b(δ2 ) = 1): C=

R sin δ2 R sin δ1 = n δ1 n tan 2 n tann δ22

.

The second parameter n is obtained by sin δ2 sin δ1 = n δ1 tan 2 tann δ22 and n ln tan

δ1 δ2 + ln sin δ2 = n ln tan + ln sin δ1 2 2

(5.12)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch which leads to n=

47

δ1 ln sin ln sin δ1 − ln sin δ2 sin δ2 = δ tan 21 ln tan δ21 − ln tan δ22 ln δ2 tan

.

(5.13)

2

The two variables C (see (5.12)) and n (see (5.13)) deliver in combination with (5.11) the radius for a conformal conic projection with two coinciding standard parallels R sin δ1 δ r= tann . n δ1 2 n tan 2

5.2

Cylinder projections

Dealing with conic projections it ´ıs appropriate to use polar coordinates like (r, ε), hence in this section we will use a Cartesian coordinate system. This system is introduced as   x = x (δ)  x (δ, α) = (5.14) y = c · R · α The parameter c denotes the scale factor of the y -axis. If there is only one coinciding standard parallel, than c = 1 for δ = π2 . In the case of c ≤ 1, i. e. the sphere and the cylinder are intersecting, we get two coinciding standard parallels δli . These are δl1 and by means of symmetry δl2 = 180◦ − δl1 . Based on (5.14) the Gaussian quantities are  dx 2 , F  = 0, G  = c2 R 2 . E = dδ Thus we can derive the principal distortions   1 dx c E G =− and b = = . (5.15) a=± E R dδ G sin δ The minus before the first principal distortion a is crucial. It indicates that x and δ are reciprocally proportional. I. e. dx /dδ < 0, or the greater x the less δ and vice versa. Since a has to be positive the differential quotient needs to be multiplied with -1. In geodesy the cylinder projections are very important. E. g. the Universal Transverse Mercator Grid-system (UTM) is based on a (transverse) cylinder projection. Also the German Gauß-Kr¨ uger coordinate system is derived from a cylinder projection. We will refer to this later.

48

5.2.1

Script ”Map Projections”

Equal area cylinder projections

The constraint V = a b = 1, standing for equal area, leads us in combination with the principal distortions to ab = −

1 dx c =1 . R dδ sin δ

After the separation of the variables and the integration x =

R cos δ + C c

is obtained. Whereas the integration constant C fixes the position of the y axis. If the parallel δ = π/2 is designed to be the coinciding standard parallel, i. e. x (δ = π/2) = 0 shall be the position of the y’-axis, than the integration constant C has to be zero. If we further take into account c = sin δl , which means coinciding standard parallels for δ1 = δl and δ2 = 180◦ − δl or as a consequence b(δl ) = c/ sin δl = 1. The resulting position vector is   cos δ x = R sin  δl . (5.16) x = y  = sin δl Rα This is a rather good example to prove, that this projection is equal area. Of course its already shown by the principal distortions a and b, hence it could also be explained by pure geometry. Looking at figure 5.1 (a) in combination with equation (5.16) we realize that the result for x is within the interval of −R/ sin δl ≤ x ≤ +R/ sin δl . On the other hand y  (see fig. 5.1 (b)) and equation (5.16)) is for all α ∈ [−π, π] → −Rπ sin δl ≤ y  ≤ Rπ sin δl . The area of this rectangle is 2 R / sin δl · 2 R π sin δl = 4 R2 π and thus the surface of a sphere with radius R. So the equal area criterion is fulfilled for the infinitesimal surrounding of an arbitrary point as well as for the finite area.

5.2.2

Conformal cylinder projections

As already shown in section 3.7.3 a projection is conformal if a = b. In combination with the principal distortions for cylinder projections (5.15) we get c 1 dx = − R dδ sin δ

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

49

+x X’

+y

δ

(b) x’= Rcos δ

(a) The cylinder surface

Figure 5.1: The calculation of the cylinder area which leads to dx = −c · R ·

dδ sin δ

.

The integration delivers x = −cR ln tan x = cR ln cot

δ +C 2

δ +C 2

and simplified .

As already explained the integration constant C fixes the y -axis. If we chose C = 0 the parallel with b(δl ) = c/ sin δl = 1 coincides with the y -axis. Hence there are also other definitions possible. Finally the position vector x of a conformal cylinder projection is   x = cR ln cot 2δ  . x = y  = cRα If c = 1 (i. e. one coinciding standard parallel) the position vector is   x = R ln cot 2δ  . x = y  = Rα

5.3

Azimuthal projections

Generally the formulas based on polar coordinates ε = α and r = r(δ)

50

Script ”Map Projections”

are applied for all for all azimuthal coordinates. Converted in Cartesian coorinates   x = r(δ) cos α  (5.17) x = y  = r(δ) sin α are derived. The derivative with respect to the parameters δ and α and the dot product deliver the Gaussian quantities  dr 2 E = , F  = 0 and G = r 2 dδ form which the principal distortions a=

1 dr R dδ

as well as b =

r(δ) R sin δ

.

(5.18)

are obtained. As shown by 5.17 the knowledge of only one parameter is sufficient. Thus knowing the radius in 5.17 means knowing the equation of the projection. This is deepened in the next sections.

5.3.1

Equal area azimuthal projection

Referring to section 3.7.3 and equation (3.49) we get in combination with (5.18) 1 dr r(δ) =1 . R dδ R sin δ This is an ordinary differential equation. By separation of the variables we get r dr = R2 sin δ dδ and after integration C 1 2 r = −R2 cos δ + or 2 2 r 2 = −2R2 cos δ + C . As the radius r for δ = 0 has to be zero we further get r 2 (δ) = 0 , i. e. C = 2R2 .

(5.19)

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Referring to (5.19) the radius for an azimuthal equal area projection is r 2 = 2R2 (1 − cos δ) regarding δ (see B.9) yields 1 − cos δ = 2 sin2 2 δ r = 2R sin . 2

5.3.2

(5.20)

Conformal azimuthal projection

By identifying the principal distortions given in (5.18) one obtains the constraints of a conformal azimuthal projection 1 dr r(δ) = R dδ R sin δ 1 1 dr = dδ and after integration r sin δ δ ln r = ln tan + ln C 2 δ r = C tan . 2 In this equation the integration constant C has to be determined. This constant can be treated as a scale factor. It is reasonable to set the scale factor for the principal point (δ = 0) to m = 1. In combination with r = m = 1 one gets R sin δ δ δ δ = R sin δ = R 2 sin cos C tan 2 2 2 δ (with δ = 0) C = 2R cos2 2 C = 2R . b =

The radius of a conformal azimuthal projection thus is r = 2R tan (see also exercise 9c in section 3.2).

δ 2

(5.21)

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Script ”Map Projections”

5.3.3

The gonomonic projcection H R M

δ

P’

r P

With this projection the center of projection is located in the center of the sphere. The radius of the polar coordinates can be derived easily to r = R tan δ and in combination with (5.18) the principal distortions are

1 dr and with R dδ dr R = whereas Figure 5.2: Gnomonic prodδ cos2 δ jection 1 1 a= as well as b= is derived. 2 cos δ cos δ a

=

The special property of the projection is, that the great circle on the sphere is mapped as a straight line and thus it the shortest distance between two points in the plane as well.

5.3.4

The orthographic projection

Observing the moon or other celestial bodies the view delivers a orthographic projection. This is the reason why celestial bodies in general are mapped with this projection.

H R

r

P’ P

δ

M

Figure 5.3: Orthografic projection

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As figure 5.3 depicts the radius r is given by r = R sin δ. The derivative with respect to δ is dr/dδ = R cos δ and thus the first principal distortion a = cos δ and the second principal distortion b=

r =1 . R sin δ

From this we get the area distortion V to V = a b = cos δ and the maximal distortion of the azimuth is sin ωm =

b−a 1 − cos δ = b+a 1 + cos δ

which leads us in combination with (B.11) to 2

sin ωm = tan

δ  2

.

,

Chapter 6 Generalized projections For this group of projections the condition F  ≡ 0 is omitted. We still distinguish the classes conic-, cylinder- and azimuthal projection, although the classification is sometimes difficult. Omitting F  ≡ 0 further means that conformal projections are no longer possible.

6.1

Pseudoconic projections

The generalized equations in polar coordinates are: r = r(δ) and ε = ε(α, δ) . The above formulas show that the parallels are concentric. The verticals however are mapped curvilinear. According to 5.1 the Cartesian coordinates are obtained with   x = rg − r(δ) cos ε(α, δ) (mit rg = R tan δg ) . (6.1) x(δ, α) = y  = r(δ) sin ε(α, δ) . Thus the Gaussian quantities can be derived with   ∂ε cos ε(α, δ) + r(δ) sin ε(α, δ) ∂δ xδ = − dr  dδ xδ = ∂ε yδ = dr sin ε(α, δ) + r(δ) cos ε(α, δ) ∂δ dδ 54

(6.2)

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and  xα

=

xα yα

∂ε = r(δ) sin ε(α, δ) ∂α ∂ε = r(δ) cos ε(α, δ) ∂α

(6.3)

to 

E =

xδ

xδ

 dr 2

∂ε dr cos2 ε(α, δ) − 2 cos ε(α, δ)r(δ) sin ε(α, δ) + dδ dδ ∂δ  ∂ε 2 +r 2 (δ) sin2 ε(α, δ) + ∂δ  dr 2 ∂ε dr + sin2 ε(α, δ) + 2 cos ε(α, δ)r(δ) sin ε(α, δ) + dδ dδ ∂δ  ∂ε 2 +r 2 (δ) cos2 ε(α, δ) ∂δ  dr 2  ∂ε 2 = + r 2 (δ) (6.4) dδ ∂δ =

as well as 

F =

xδ

·

xα

∂ε  ∂ε dr cos ε(α, δ) + r(δ) sin ε(α, δ) r(δ) sin ε(α, δ) = − dδ ∂δ ∂α  dr ∂ε  ∂ε + sin ε(α, δ) + r(δ) cos ε(α, δ) r(δ) cos ε(α, δ) dδ ∂α ∂α ∂ε ∂ε ∂ε ∂ε = r 2 (δ) sin2 ε(α, δ) + r 2 (δ) cos2 ε(α, δ) ∂δ ∂α ∂δ ∂α ∂ε ∂ε (6.5) = r 2 (δ) ∂δ ∂α 

together with  ∂ε 2  ∂ε 2 G = xα · xα = r 2 (δ) sin2 ε(α, δ) + r 2 (δ) cos2 ε(α, δ) ∂α ∂α  ∂ε 2 = r 2 (δ) . (6.6) ∂α

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Script ”Map Projections”

The intersection angle of the parameter curves according to 2.3 is √ E  G − F 2  √ √ sin τ = E  G    2   2  2  2 2 dr ∂ε 2 ∂ε 2 ∂ε 4 ∂ε r + r − r dδ ∂δ ∂α ∂δ ∂α   =     dr dδ

=   2 dr dδ

dr dδ

+

r2

2

+ r2

∂ε ∂δ

2

r2

∂ε ∂α

2

(6.7)

 2 ∂ε ∂δ

Thus the scale distortions h and k along the parameter curves are $   %  √ % dr + r 2 ∂ε 2  & dδ ∂δ E 1 dr dδ = as well as h = √ = 2 R R sin τ E $ %  2 √ % r 2 ∂ε ∂ε  & ∂α G 1 r ∂α . = k = √ = R sin δ R2 sin2 δ G

(6.8)

(6.9)

The distorion of area V is derived as dr ∂ε r sin τ  V =a·b=h·k = dδ2 ∂α sin τ R sin δ

(6.10)

and the principal distortions are a±b=

√

h2 + k 2 ± 2V

.

If we want to fulfill the equal area criterion based on the above formulas we get dr ∂ε r (6.11) V = dδ2 ∂α = 1 R sin δ and this expression can be simplified to R2 sin δ ∂ε = dr ∂α r dδ

.

(6.12)

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Separating the variables of this differential equation combined with the integration yields R2 sin δ ε= α+C . (6.13) dr r dδ A quite obvious choice for the integration constant is C = 0. As a consequence for α = 0 also ε = 0 and thus the according vertical is a straight line.

Figure 6.1: Bonne projection As a further condition we postulate equal area and no scale distortion along the parallels. The latter means k = 1 which leads us to ∂ε r ∂α =1 R sin δ and together with the relationship 6.12 we get

(6.14)

2

δ r R r sin dr dδ

=1⇒

R sin δ The integration yields the radius to

dr =R . dδ

r =R·δ+C

.

(6.15)

(6.16)

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Script ”Map Projections”

It is r(δ = 0) = C. This shows that C determines the pol of the projection. Hence the pole depends on the standard parallel because rg = R tan δg . Thus C + Rδg = rg ⇒ C = rg − Rδg = R(tan δg − δg ) and furthermore r = Rδ + R(tan δg − δg ) . Now ε can also be determined explicitly. It is ε=

sin δ R2 sin δ α= α . R · R(δ + tan δg − δg ) δ + tan δg − δg

(6.17)

This projection (see figure 6.1) is named after Rigobert Bonne (1727 - 1795) who devellopped it. Sometimes it is also called Bonne equal area pseudoconic projection.

Figure 6.2: Stab-Werner projection In the specific case for δg = 0 we get from 6.17 the relationship ε=

sin δ α and r = R · δ δ

.

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

59

This projection is called Stab-Werner projection or Stab-Werner heart-shaped pseudoconic projection. Because the standard parallel coincides with the pole this projection can also be seen as a pseudoazimuthal projection.

6.2

Pseudocylindrical projections

The general equation for these projections are  

x (δ, α) =

x y

= x (δ) = y  (α, δ)

(6.18)

Thus on pseudocylindrical projections the shape of the parallels are straight lines and the meridians (verticals) are curves. The symmetry axis is the central meridian. The directional derivatives are  xδ

=

and

 xα

=

xδ yδ

xα yα

= =

dx dδ ∂y  ∂δ

= 0  = ∂y ∂α

.

From the above equations the Gaussian quantities are derived. These are E  = xδ · xδ = F G

 dx 2

+

dδ ∂y  ∂y  = xδ · xα = · ∂δ ∂α  ∂y  2 = xα · xα = ∂α

 ∂y  2 ∂δ

(6.19) (6.20) (6.21)

As already discussed in section 5.2 the differential quotient dx /dδ < 0 is also negative. Thus we need a negative sign. The intersection angle of the

60

Script ”Map Projections”

parameter curves therefore is √ E  G − F 2  √ √ sin τ = E  G    dx dδ

=

= − 

2    2   2   2 ∂y + − ∂y · ∂y ∂α ∂δ ∂α       2 2 2 ∂y  ∂y  dx + · dδ ∂δ ∂α 2

dx dδ

dx dδ

2

+

∂y  ∂δ



∂y  ∂δ

2

.

(6.22)

The distortions along the parametric curves h and k are given by $ %  2   2  % dx ∂y E  & dδ + ∂δ = , h= E R2 which leads us together with 6.22 to 

dx 1 h = − · dδ  R sin τ

and thus we obtain the first distortion parameter. The other is given by $ %   2  % ∂y ∂y  G & ∂α = k= = ∂α G R sin δ R2 sin2 δ and the distortion of area can be written as 

∂y 

1 dx sin τ  dδ ∂α =− sin τ  V =a·b = h·k· sin τ R sin τ  R · sin δ dx ∂y  1 · · . = − 2 R · sin δ dδ ∂α

(6.23)

Thus the principal distortions are √ h2 + k 2 ± 2V a±b =   ∂y  2 1 dx ∂y  1 1 1  dx 2 (6.24) . ∓ 2 · + · = R dδ ∂α sinδ dδ ∂α sin2 τ  sin2 δ

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Replacing sin τ  by 6.22 delivers after some deductions the formulas for the principal distortions  1  dx ∂y  1 2  ∂y  2 a±b = ∓ + . (6.25) R dδ ∂α sin δ ∂δ

6.2.1

Equal area with equidistant parallels

The condition for equidistant parallels is ∂y  1 k=1= · ∂α R · sin δ

.

The separation of the variables and integration deliver for the second component of the position vector y  y  = R sin δ · α + C

.

The only reasonable solution for the integration constant is C = 0. Because only this definition delivers equidistant parallels based on a radius of R · sin δ which already corresponds with the radius of the parallels. Therefore we obtain

y  = R · sin δ · α

and thus

∂y  = R · sin δ ∂α We impose the equal area condition

.

(6.26)

 dx  R · sin δ V =1=− · 2 dδ R · sin δ from which we get after some deductions x = −R · δ + C

.

(6.27)

Presuming that the y  -axis coincides with the equator, i. e. x (δ = π/2) = 0, we get as a result for the integration constant C =R·

π 2

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Script ”Map Projections”

and together with 6.27 x = −R · δ + R ·

π  π =R· −δ 2 2

is obtained. The resulting position vector is    π  x = R · − δ 2 x =  y = R · sin δ · α

.

(6.28)

In the literature this projection is called Mercator-Sanson projection (see exercise 2 and figure D.20), Sanson-Flamsteed projection or sinusoidal projection.

6.2.2

Equal area with elliptical meridians

Deducing this projection we first impose that the surface of the unit hemisphere is mapped on a circle with radius r. I. g. √ OHK = 2π und FK = r 2 π from which we derive r = 2 . This is the radius of the circle depicted in figure 6.3 and according to the sketch at the same time the semi minor axis of the ellipsoid denoted further with b. Imposing the equal area condition the area of the surface of the unit sphere must be the same as the area of the ellipse. Thus √ a · b · π = 4 · π and in combination with b = 2 the semi major axis of the ellipse is a = 2·

√

2 .

From figure 6.3 the relationship x (ψ) =

√

2 · R · cos ψ

(6.29)

can be derived. However we have to take into account, that no longer the unit sphere, but the sphere with the radius R has to be mapped. Considering

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x B

C ψ

A

D

y

Figure 6.3: The derivation of the projection of Mollweide

6.18 the first component of the position vector x = f (δ) is not yet solved, because it depends on δ and not on ψ. From geometry we derive √ y  = 2 2R · sin ψ

.

The above expression in combination with the equation 6.29 obviously describe the shape of the ellipse. Considering α and the fact that α ∈ [−π, π] we get √ 2 2 y  (ψ, α) = · R · α · sin ψ . π Thus we got the components of the position vector: √   x = 2 · R · cos ψ √ x (ψ, α) = (6.30) 2 2  · R · α · sin ψ y = π As already mentioned the above formula delivers x (ψ, α), however this not what we want. Hence we need a relationship between ψ and δ. This is emphasized by 6.18. If we take the equal area criterion 6.23 into account we get V = h · k sin τ  = −

dx ∂y  1 · · =1 . R2 · sin δ dδ ∂α

(6.31)

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Script ”Map Projections”

From 6.30 can be derived √ dψ dx = − 2 · R · sin ψ dδ dδ

and

√ ∂y  2 2R = · sin ψ ∂α π

,

which leads us together with 6.31 to equation √ √ 2 2 2 · R · sin ψ dψ · · sin ψ · R dδ π =1 . 2 R · sin δ This equation encompasses the differential relationship between the two parameters ψ und δ. Separation of the variables leads to π sin2 ψdψ = · sin δdδ , 4 and after integration we get π ψ 1 − sin(2ψ) + C = − cos δ 2 2 4

.

(6.32)

Imposing ψ = 0 if δ = 0 yields for the integration constant π . C=− 4 In combination with 6.32 one obtains after some transformations ψ 1 π (1 − cos δ) = − sin(2ψ) 4 2 4 and together with sin2

1 δ = (1 − cos(δ)) (see (B.9)) 2 2

finally after the multiplication with 4 2ψ − sin(2ψ) = 2π sin2

δ 2

(6.33)

is yielded. This equation is transcendent and can only be solved in an iterative way. One of the best approaches to so is the Newton-method 1 . Equation 6.33 delivers a relationship between ψ and δ. Thus the azimuthal coordinates can be calculated and finally also the position vector on the image plane. Further 6.18 is fulfilled. 1

Sometimes also called Newton-Raphson method

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6.2.3

65

Equal area with elliptical meridians and pole lines

The position vector of this projection is given by ⎧ ⎨ x = √ 2Rπ · cos ψ π(4+π)  x = 2R·α  √ y = · (1 + sin ψ) ⎩

(6.34)

π(4+π)

in combination with the transcendent equation for ψ π − (4 + π) cos δ = 2ψ − 4 cos ψ − sin 2ψ

6.2.4

.

With equidistant equator and equidistant central meridian N P(x’,y’)

r(α)

p

φ e(α)

The images of the circles of latitude have to be equidistant parallels to the y  -axis and further the image of the central meridian has to be equidistant as well. This delivers for x = f (δ) the relationship  π −δ . (6.35) x = R 2

In addition the images of the meridians should be circles through the pole. The centers of the circles thus have S  all the coordinate xM = 0 and yM = e(α). The equation of the circles are Figure 6.4: Sketch to explain the −¯ thus projection of Apian x2 + (y  + e¯)2 = r¯2

.

For the equator it is q =R·α .

(6.36)

Further on we can deduce x (0) = p = R

π 2

(6.37)

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Script ”Map Projections”

and r¯ = e¯ + q It is

.

e¯2 + p2 = r 2 = (¯ e + q)2 = e¯2 + 2¯ eq + q 2

,

from which we get the relationship p2 = 2¯ eq + q 2

or p2 − q 2 = 2¯ eq

Resolved to e¯ finally yields e¯ =

.

p2 − q 2 2q

and after insertion of (6.37) and (6.36) one obtains R(π 2 − 4α2 ) 8α

e¯ = Thus it is

r¯ = e¯ + q =

.

(6.38)

R(π 2 − 4α2 ) +R·α 8α

and summarized

R(π 2 + 4α2 ) . 8α With |α| = π/2 we get for the image of the meridian bordering the hemisphere the realtionship π  π  π e¯ = 0 und r¯ =R· . 2 2 2 r¯ =

The angle φ is

p e¯ and applying the relationship 6.37 as well as 6.38 we get the equation tan φ =

tan φ =

π2

4 · πα − 4 · α2

.

(6.39)

In combination with (B.11) we get for tan φ/2 tan

φ 2

=

p r¯

1+

e¯ r

=

2·α p = r¯ + e¯ π

.

(6.40)

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The quantities e¯ and r¯ can be transformed regarding (6.38) as well as (6.39) to R · cot φ · 4 · π · α R · cot φ · π e¯ = = (6.41) 8·α 2 and subsequently with (6.40) can be written as     2 φ 2 φ     cos 2 + sin 2 R φ R φ R     r¯ = · π · cot . + · tan ·π = ·π 4 2 4 2 4 sin φ · cos φ 2

2

We simplify the relationship for r¯ by applying (B.4). Finally one obtains r¯ =

R·π 2 · sin φ

.

(6.42)

According to Pythagoras it is for an arbitrary point P (y  + e¯)2 + x2 = r¯2 or resolved for the first term (6.43) (y  + e¯)2 = r¯2 − x2     2 2 2 2 2 2 R ·π R ·π 2 π 2 π 2 − δ − πδ + δ − R − R = = 2 4 4 · sin2 φ 4 · sin2 φ     2 2 R ·π 2δ 2 · 1 − sin2 φ 1 − = . (6.44) 2 π 4 · sin φ Regarding (6.41) we get for the root Rπ R·π cot φ = y + e¯ = y + 2 2 · sin φ 





 2δ 2 1 − sin2 φ · 1 − π

.

(6.45)

Differentiating the relationships for x and y  with respect to δ we get regarding the aspects discussed in section 6.2 for the first parameter dx = −R . dδ The second derivative is based on the implicit relationship given by equation (6.43). One obtains 2 · R2 · π 2  2 · δ 2 ∂ y 2(y  + e¯) = − 1− ·− ∂δ 4 π π

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Script ”Map Projections”

further reduction of the fraction delivers   2 π R 2 −δ ∂ y = ∂δ y  + e¯

(6.46)

a simplified formula. Inserting the above results in (6.22) delivers sin τ  = 

1

 1+

R2



π 2

−δ

2

(y  + e¯)2



which leads to

 2δ 2 1 − sin2 φ 1 − π after some transformations. Inserted in equation (6.45) yields sin τ  =

y  + e¯ =

(6.47)

R · π sin τ  2 · sin φ

and resolved to y  , considering for e¯ the relationship (6.41), one finally obtains y =

R · π(sin τ  − cos φ) 2 · sin φ

.

Thus we found the position vector for the image plane. Collecting all we get    π  x = R − δ 2 (6.48) x = R·π(sin τ  −cos φ)  y = 2·sin φ in combination with the relationship for φ and τ    4·π·α 2 · δ 2  2 tan φ = 2 und sin τ = 1 − sin φ 1 − π − 4 · α2 π

.

Chapter 7 Special projections and quality aspects In this chapter different aspects of world maps are discussed. The first section is dealing with projections encompassing specific properties. In the ensuing sections quality aspects are emphasized. The term quality is meant in a narrower sense of the lifelike representation of the globe. This actually presumes mathematical tools for the decision-making process. Different approaches are introduced and discussed. According to a proposal introduced by Lee projections which are neither conformal nor equal area are called aphylactic – some of them were already presented before although the term ”‘aphylactic”’ was not mentioned. In this script only two of this category are presented. Actually there are much more such projections, e. g.: • Trapezoidal • van der Grinten • Gnomonic (see fig. 5.2) • Orthographic (see fig. 5.3) • Airy • ... 69

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Script ”Map Projections”

The Robinson projection is used to explain different evaluation tools to summarize the quality or the result of a projection. Therefor the different implementation suggestions are compared with each other to decide which of the solutions delivers the best results according to the quality criteria.

7.1 7.1.1

Aphylactic small scale maps The Robinson projection

In the past the Robinson-Projection was chosen by seven different cartographic or geographic organizations amongst three others for the representation of world maps. This kind of projection however is not comparable to the above mentioned - e. g. the projections of chapter 5 - because it lacks a rigorous mathematical concept.

Figure 7.1: Robinson projection As already mentioned Arthur H. Robinson published in his paper in 1974 only a few instructions for the construction of this projection. Evolving the projection he started with some fundamental issues:

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Table 7.1: Distances/Lengths between/of the parallels Latitude Distance between the parallels Length of the parallels 90◦ 1.0000 0.5322 ◦ 0.9761 0.5722 85 ◦ 80 0.9394 0.6213 0.8936 0.6732 75◦ ◦ 70 0.8435 0.7186 65◦ 0.7903 0.7597 ◦ 0.7346 0.7986 60 ◦ 55 0.6769 0.8350 0.6176 0.8679 50◦ ◦ 0.5571 0.8962 45 ◦ 40 0.4958 0.9216 0.4340 0.9427 35◦ ◦ 0.3720 0.9600 30 25◦ 0.3100 0.9730 ◦ 0.2480 0.9822 20 ◦ 15 0.1860 0.9900 0.1240 0.9954 10◦ ◦ 0.0620 0.9986 5 0◦ 0.0000 1.0000 • The projection should fit wall maps and other representation of the whole surface of the earth. • The result of the mapping should be continuous. • Angle as well as the distortion of the area has to be minimal. Robinson’s conclusion were as follows: • The projection cannot be equal area. • Cylinder projections are not fitting the problem. • It is impossible to map the poles as points. • ...

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Script ”Map Projections”

Summarizing the above topics Robinson concluded further that the projection he was looking for can only be found by systematic “trial and error”. Based on this idea he deduced a set of rather simple construction principles: • The length of the equator is 0.8487 times as long as the circumference of the sphere sharing the same surface as the earth. • The prime meridian is 0.5072 time as long as the equator of the projection. • On all parallels the meridians are equidistant. • The parallels are straight lines and parallel to the equator. Robinson compiled a table — by “trial and error” as already explained — encompassing the distances between the parallels. The numerical result of this trial is shown in table 7.1 whereas the visual result is depicted in figure 7.1. Several mathematical cartographers tried to approximate the Robinson Projection by different ways. Predominantly four solutions are discussed: • Canters and Decleir presented their solution in 1989 • Snyder published his approach one year later • Beineke followed again one year later with his approach • In 1994 Bretterbauer presented a new algorithm A visual comparison between the approach of Bretterbauer and Canters & Decleir is depicted in figure 7.2 All authors tried to find an optimal solution. Whereas some stick very close to the proposals of Robinson others calculated algorithms fitting (e. g. least squares) the support points. Examples: The general transformation formulas are: x = f (ϕ) y = f (ϕ, λ)

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Figure 7.2: Bretterbauer (black) and Canters & Decleir (red) 1. Approach (Beineke) x = a · ϕ + b · s· | ϕ |c

(7.1)

y = (d + e · ϕ2 + f · ϕ4 + g · ϕ6 ) ·

λ π

(7.2)

with  1 for ϕ ≥ 0 s := sig (ϕ) = −1 for ϕ < 0 and a =0.96047 d =2.6666

b = − 0.00857 e = − 0.367

c =6.41 f = − 0.150

g =0.0379

2. Approach (Bretterbauer) Bretterbauer suggested a solution using splines to approximate the images of the meridians. For convenience a new parameter set with modified coeffi-

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Script ”Map Projections”

cients A∗ and B ∗ is introduced. The transformation formulas are then y = R · A∗ · λ x = R · B∗ The new coefficients A∗ and B ∗ are depicted in table 7.1.1. Table 7.2: The coefficients computed by Bretterbauer i ϕ(i) [◦ ] A∗ B∗ 0 0 0.84870000 0.00000000 1 5 0.84751182 0.08384260 2 10 0.84479598 0.16768520 3 15 0.84021300 0.25152780 4 20 0.83359314 0.33537040 5 25 0.82578510 0.41921300 6 30 0.81475200 0.50305560 7 35 0.80006949 0.58689820 8 40 0.78216192 0.67047034 9 45 0.76060494 0.75336633 10 50 0.73658673 0.83518048 11 55 0.70866450 0.91537187 12 60 0.67777182 0.99339958 13 65 0.64475739 1.06872269 14 70 0.60987582 1.14066505 15 75 0.57134484 1.20841528 16 80 0.52729731 1.27035062 17 85 0.48562614 1.31998003 18 90 0.45167814 1.35230000 3. Approach (Canters & Decleir) This approach is based on a polynomial approximation. The coefficients of the polynomial are derived by the least squares method. As a consequence the polynomials don’t exactly fit teh support points. Hence Canters & Decleir claim that the results are sufficient for wall maps. The formulas are as follows

 y =R · Λ A0 + A2 ϕ2 + A4 ϕ4 

x =R A1 ϕ + A3 ϕ3 + A5 ϕ5

(7.3) (7.4)

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in combination with coefficients A2 = −0.1450 A3 = −0.0013

A0 = 0.8507 A1 = 0.9642

A4 = −0.0104 A5 = −0.0129

4. Approach (Ipbuker) Table 7.3: The coefficients i ϕ(i) [◦ ] 0 0 1 5 2 10 3 15 4 20 5 25 6 30 7 35 8 40 9 45 10 50 11 55 12 60 13 65 14 70 15 75 16 80 17 85 18 90

pj and qj according to the approach of Ipbuker p q 0.40711579454 0.91083562255 -0.00875326537 0.00000589975 0.01069796348 0.00000564852 -0.01167039606 0.00000557909 -0.00680782592 0.00000555879 0.01847822803 -0.00000001291 -0.02090931959 0.00000546138 -0.01847842619 -0.00154708482 0.02090971277 -0.00387351841 -0.01410147990 0.00619324913 -0.02236858853 0.00930492848 0.01701955610 -0.01239340212 0.01215649454 0.01549814705 -0.01069792545 0.01937169560 -0.02090967766 -0.02401844414 0.03160740722 0.03331171624 0.01361549135 0.07051393824 0.04425022432 -0.09917388904 0.60843116534 0.24527101656

The bi-cubic interpolation method of Ipuker should be presented as a third example. The formulas are given by yi =R · Λi

18 '

pj |5j − ϕi |

(7.5)

j=0

xi =R

18 ' j=0

qj |5j − ϕi |

(7.6)

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Script ”Map Projections”

with the coefficients for the parameters pj and qj as depicted in table 7.3

Pi (ϕ, λ) P1 (30◦, 30◦) P1 (30◦, 60◦) P1 (30◦, 180◦)

Table 7.4: Comparison of three points Beinecke Canters & Decleir y = 0.42592 x = 0.50277 y = 0.42420 x = 0.50416 y = 0.71125 x = 0.99428 y = 0.71124 x = 0.99197 y = 1.41718 x = 1.35378 y = 1.34966 x = 1.38616

A numerical comparison between Beinecke and Canters & Decleir is given in table 7.4. There the coordinates of three points are depicted to explain numerically the difference between the two solutions.

7.1.2

Chamberlins trimetric projection

This is an easy to understand projection developed by Wellman Chamberlin in 1946. The projection is based on only a few principles depicted below: • The area of the datum surface to be mapped has to be enclosed by a spherical triangle. • Thus we get the three legs each representing a part of a great circle. These legs now are taken as – with respect to the mapping scale – the legs of a plane triangle. • All other points on the datum surface are mapped to the plane in a similar way. First the great circles of the datum surface connecting each vertex of the triangle with the point to be mapped are determined. • Afterwards the edge defined by the point and the vertex of the triangle is mapped to the image surface too (also with respect to the scale). As a result we don’t get a point but an area representing the mapped point. • The centroid of the above mentioned are could be treated as the representation of the point on the datum surface. However, Chamberlin didn’t define the algorithm for the final point determination. Chamberlin’s projection has two disadvantages being crucial. First

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77

• it is not possible to generate a world map based on the above principles. Because of the excess of the spherical triangles, their plane images are not continuous. • And second as already indicated afore Chamberlin didn’t define a rigorous transformation algorithm. Thus up to a certain level the mapping of the points of the datum surface is arbitrary which means not clearly determined. Hence some scientists derived interpolation algorithms to fix this disadvantage. Although this projection didn’t encompass a strict mathematical concept the National Geographic Society (NGS) used it for several mapping projects.

Figure 7.3: Graticule of the trimetric projection of W. Chamberlin

7.2

Optimal projections

In this chapter we try to define and find so called optimal projections by applying restrictions concerning the distortions. Thus we get optimized projections with respect to specific parameters. First we will limit our view on local distortion quantities such as the principal distortions.

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Script ”Map Projections”

7.2.1

Local distortion quantities

In this context the following quantities are of interest: 1. Airy 1 2A = ((a − 1)2 + (b − 1)2 ) 2 2. Airy-Kavrajski 1 2AK = (ln2 a + ln2 b) 2 3. L¨obell 2L =

 a2 − b2 2 2ab

4. Fiorini 2F

2

= (a b − 1) +

5. Jordan 2J

1 = 2π

 0

6. Jordan-Kavrajski 2JK 7. Klingatsch 2Kl =

2π

1 = 2π

a b

−1

2

(m − 1)2 dβ



2π

ln m2 dβ

0

    Pω ab − 1 + PV (a · b − 1)2 Pω + PV

Where Pω is the weight for the maximal distortion of the angle and PV is the weight depending on the area distortion. Most of the above local distortion criteria could be explained geometrically. This should be deduced exemplary by the criterion of Airy which is based on the ellipse of distortion treated as the affine projection of the circle. The differences of the circle and the ellipse points form the basis for the optimization procedure. These difference is shown in figure 7.4.

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Figure 7.4: Geometric explanation of the criterion of Airy The local distortion quantities are all based on the ellipsoid of distortion. Thus they simply represent the immediate vicinity of a specific point. Which means they could be used to derive optimal projections with respect to specific points. This is what should be demonstrated in exercise 10. 7.2.1.1

The image of the finite circle

If we consider the Robinson projection – in its pure definition – we got no mapping functions to calculate the indicatrix and it is the same with the trimetric projection. In some other cases (e. g. Eckert II, see figure D.18 together with formula A.4) the mapping functions are so complicated that it is very difficult or almost impossible to get the elements – which means extreme values – of the indicatrix.

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Script ”Map Projections”

In all these cases we can use a finite circle on the surface of the sphere and project a finite number of points of the circumference to the plane analyzing the distortions now by comparison the small circle and its affine image. So we have zu find the extrem values of the radius vectors on the image plane to get the elements of the indicatrix. There is actually a difference between the result derived by the differential calculus and Figure 7.5: The small circle the outcome outlined afore, hence the deviation can be ignored. Thus we use the image of the finite circle as a substitute for the indicatrix. Further we will use the term indicatrix for both images of the circles, the one we got by differential calculus and the one we got point by point.

7.2.2

Global distortion quantities

Unlike the local distortion quantities, the global distortion quantities are suitable to compare map projections. The overall distortion is taken into account, not only a specific surrounding. As a consequence we are able to decide, if a specific projection fits the purposes or if its better than another. This is why these quantities are mentioned in this section entitled with ”‘Optimal projections”’. Thus the term optimal is treated in a more general way so the projection need not be the result of an optimization process in the strict sense. The widely known distortion quantities are discussed below. 7.2.2.1

The concept of Peters

In this context the suggestion of Peters has to be mentioned. He refined an idea of Tobler who suggested to analyze the distortion of a large set of

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randomly placed triangles on the datum surface.

Figure 7.6: Peters’ approach Peters implemented this approach into a computer program and thus compared the difference in length between 30 000 lines. Therefor he calculated 60 000 points randomly placed on the sphere connected each two points with a Orthodrome and compared the length of the 30 000 great circles with their projected images. From this comparison he derived a new quantity called Ep which is defined as follows n 1 ' |Si − Si | (7.7) EP = n i=1 |Si + Si |    Si  n 1 − Si  1'   (7.8) = n i=1 1 + Si  Si with Si : length of the orthodrome (sphere) Si : length of the straight line (image) n : number of the lines .

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Script ”Map Projections”

EP can be used to rank maps or to compare maps regarding especially the distortion of length. Peters however never initiated such a comparison.

7.2.2.2

Canters’ scattered circles

A very interesting method to describe the quality of a projection is presented by Canters. According to his proposal a map can be analyzed by comparing the randomly scattered circles on the datum surface with their images of the plane. Thereby the maximum radius of such a circle can be 30◦ . The resulting deformation of the finite circles EA is derived form 16 radius vectors thus enclosing 22,5 degrees between each other. Based on the transformation of these 16 radius vectors – which are great circles on the datum surface – the area distortion is computed. Canters offers the following formula n 1 ' |Ai − Ai | EA = . n i=1 |Ai + Ai | The quantities Ai and Ai denote the areas of the datum surface and the plane respectively. However, the shape of the projected circle has to be taken into account too. Therefore the shape index according to Boyce & Clark has to be calculated (see B.2). This is done by   n    ' 100 r   ES =   (16i  · 100 −   j=1 rj n i=1 In the above formula the quantity ri denotes the radius vector of the projected circle on the plane. The result is called the shape index of the circle and denotes the derivation of the resulting plane figure from an ideal circle. The shape index for a circle is 0, whereas the shape index of all other geometric forms are greater than 0. The next step conforms to a normalization. To get rid of the unequal units the quantities EA and ES are mapped on the interval [0,1]. This is expressed

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83

Figure 7.7: Canters’ scattered circles

by 1 (KA − KAmin ) and KAmax − KAmin 1 = (ES − ESmin ) in combination with ESmax − ESmin 1 + EA = 1 − EA

EA,c = ES,c KA

The above formulas however require minimal and maximal values for the quantities KA and ES standing for the upper and the lower bound of the interval. For the area distortion this can be achieved quite easy because the minimal value is 1. Hence, the maximum distortion is difficult to find, because there is no theoretical limit. Canters’ analysis of different projections delivered the following upper and

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Script ”Map Projections”

lower bounds: ESm in ESm ax KAm in KAm ax

= = = =

0.0356 0.1760 1 1.8211

The pure addition of the two values ES,c and EA,c for any arbitrary projection delivers its characteristic total distortion value E. Expressed as a formula we get E = EA,c + ES,c . As depicted in figure 7.7 the placement of the circles is predominantly limited to the landmasses. Nonetheless some of the circles also cover parts of the worlds water areas. This is due to the preconditions of the randomly placed circles. These conditions are: • The center of a circle has to be placed on a continental area. • At least twelve out of sixteen endpoints of the above mentioned radius vectors have to be situated on the landmasses. Conversely four of them can be situated on water areas. 7.2.2.3

Capek and the limits

Capec also presented a method to describe the quality of small scale maps. Unlike the hitherto mentioned approaches the ellipse of distortion is focused. Based on this ellipse from which we easily can derive the angular as well as the areal distortion (see section 3.3 and 3.2) Capek simply defined maximum values for both. He called the combination of the two quantities the distortion characterization Q. The limits for the deviations are more or less arbitrary, however, Capek found out, that a maximum angular distortion of 40◦ and a maximum distortion of the area of 1.5 are fairly good limits for the assessment of cartographic projections. In his analysis of different projections, Capek used the afore mentioned limits as extreme values and evaluated the percentage of the projection surface

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which is in compliance with Q. In other words he calculated the percentage of the area where the angular and areal distortion are less than the prescribed limits. Table 7.5 gives an overview about the ranking of several small scale maps. Table 7.5: The ranking of different projections based on Q Projection Q [%] Robinson 82.6 Eckert IV 82.5 Kavraiskii VII 82 Wagner VI 79.5 Mollweide 70.4 Aitoff 66.6 Some of the above mentioned projections can be found in appendix D.7. 7.2.2.4

Approaches based on the distortion ellipse

First the proposal of Tschebyscheff should be outlined in this section. The principle is based on the mean value theorem and can be expressed by  1 ε2 dF. εm = F F i Whereas F is a specific area and εi is one of the local distortion quantities mentioned in section 7.2.1. For practical purposes we replace the integral by the sum and thus we we get 1' 2 εm = ε dF. k i=1 i k

Thus we sum up the local distortions of small areas which are treated as a representative of the point and its vicinity.

Chapter 8 Further concepts – isometric coordinates 8.1

Isometric surface parameters

In this section the parameters forming a quadratic net are focused. Hence the Gaussian quantities are E(u, v) = G(u, v) = μ2 (u, v) ,

F (u, v) = 0, and the line element becomes

ds2 = μ2 (u, v)(d u2 + d v2 ) . Assuming the same for the projection surface we get ds2 = μ2 (u, v)(d u2 + d v2 ) . representing the line element. Thus the line scale λ2 =

ds2 μ2 (u, v) = ds2 μ2 (u, v)

only depends on the parameters u and v. Supposing the plane as the projection surface we obtain (E = G = 1) ds2 = dx2 + dy 2

(x = u and y  = v) . 86

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87

Together with μ2 = 1 we get as a result for the square of the line element λ2 =

1 μ2 (u, v)

.

Now λ depends only on the both parameters u and v). To get plane conformal projections (F  = 0 and E  = G = 0) the following equations have to be fulfilled xu xv + yu yv = 0 und 2 2 x2 = x2 u + yu v + yv

.

(8.1) (8.2)

Solving (8.1) for xv and put it in (8.2) we obtain xu = ±yv und xv = ±yu . These equations are identical withe the Cauchy-Riemann differential equations, denoting the relationship of partial derivatives and the real and imaginary part of complex functions. This can be expressed by x + iy  = f (u + iv) . Thereby the the function f (u + iv) is regular and analytic which means it is continuous, can be differentiated and further the function fulfills the CauchyRiemann differential equations. For practical purposes these functions can be used to derive conformal projections. Therefore we only need to create complex quantities out of the isometric parameters. This can be done by w = u + iv

on the datum surface

and z = x + iy  (plane) or z = u + iv (general surface) as well z = f (w) respectively.

88

8.2

Script ”Map Projections”

Isometric parameters on the sphere

The line element can be expressed in geographic coordinates by ds2 = R2 db2 + R2 cos2 b dl2

.

Factoring out the term R2 cot cos2 b delivers  db2  2 2 2 2 + dl ds = R cos b cos2 b and the substitution of db2 dq 2 = cos2 b leads to the isometric form ds2 = R2 cos2 b(dq 2 + dl2 ) .

(a) Geographic Coord.

(b) Isometric Coord.

Figure 8.1: The different graticules

The isometric latitude can be derived by integration of  db which delivers q = cos b π b  + = ln tan . 4 2

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89

The isometric longitude remains unchanged. Figure 8.1 depicts the geographic as well as the isometric parameter system on the sphere. It clearly shows that the isometric parameter system gets closer towards the poles.

8.3 8.3.1

Conformal projection of the sphere Mercator projection

Based on the isometric parameter system for the sphere derived in the section above given by w = q + il and the isometric coordinates of the plane z = x + iy 

.

The regular analytical function z = w maps the sphere to the plane. The −180

−120

−60

+60

+120

+180 +80

+60

+40 +20 0 −20 −40

−60

−80

0

Figure 8.2: The Mercator projection separation of the real and the imaginary part delivers x = q = ln tan y



= l

.

π 4

+

b and 2

90

Script ”Map Projections”

Assuming an unit sphere (R = 1) the line scale λ can be deduced from λ=

1 cos b

.

Due to lim λ = −∞

lim λ = +∞ und

b→ π2

b→− π2

the poles can not be mapped by this projection.

8.3.2

Stereographic projection

Relating to the regular analystical function z = −ke−z (representation according to Euler) the substitution of z = w delvivers z¯ = −ke−w = −ke−(q+il) = −ke−ln tan( 4 + 2 ) e−il π

b

.

(8.3)

In combination with eiπ = cos π + i sin π = −1 we get π

π b b + − r = k cot = k tan and 4 2 4 2 Φ = π−l . Based on these formulas we conclude 1. A variation concerning the latitude affects only the radius. As a consequence the parallels are concentric circles. 2. If the latitude l remains unchanged the quantity Φ remains unchanged too. Thus the meridians are straight lines. 3. The line scale equals

 cot

λ=k

π 4

+

cos b

b 2

 =

2 sin2

k 

π 4

+

b 2



and further b = π/2 ⇒ r = 0. Postulating λ(π/2) = 1 which means no distortion at the north pole we get k = 2.

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Finally we derive π

b  i(π−l) − z = 2 tan e 4 2

(again in accordance with the notation of Euler).

The transformation to Cartesian coordinates delivers ⎧     ⎨ x = 2 tan π − b cos(π − l) = −2 tan π − b cos(l)  4 2 4 2  x = ⎩ y  = 2 tan π − b sin(π − l) = 2 tan π4 − 2b sin(l) 4 2 (8.4) Figure 8.3 depicts the graticule of the Stereographic projection. x’ −135°

135°

y’

30° −45°

45° 0°

Figure 8.3: Stereographic projection

8.3.3

Projection of the sphere on a sector

In this section the plane z¯ is mapped conformal on the sector z˜ or expressed by a formula z˜ = z¯n , where as the factor n is an arbitrary real number within the interval n ∈ [0, 1] . Further we need a rotation in oder to get the prime meridian to the south. Thus it is z = z¯n eiπ(1−n) .

92

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In combination with 8.3 the equation changes to ˜

z˜ = r˜eiΦ = k n cotn

π 4

+

b  i(π−nl) e 2

.

The substitution of k¯ = k n leads to the transformation formulas x’ −180°

180°

y’

30° −60°

60° 0°

Figure 8.4: Projection on a sector

r˜ = k¯ cotn

π 4

+

π b b ¯ = k tann − 2 4 2

and ˜ = π − nl. Φ Based on k and n further restrictions can be determined. The graticule in figure 8.4 is depicted for n = 0.75 and b = 50◦ as a standard parallel.

Appendix A Exercises Exercise 1 The datum surface ⎧ ⎨ x = R cos u cos v x(u, v) = y = R cos u sin v ⎩ z = R sin u

(A.1)

is mapped on the image surface ⎧  ⎨ x y x (u , v  ) = ⎩  z

= r cos v  = r sin v  = u

(A.2)

based on the below formulas v u r

= v = R tan u = R

(A.3)

(R = 6370 km). Further two points are given: K (λK = 8◦ east, ϕK = 49◦ north) and M (λM = 38◦ east, ϕM = 56◦ north). a) Calculate the position vector x (u, v)? b) How can this mapping be interpreted geometrically? 93

94

Script ”Map Projections” c) Calculate the Cartesian coordinates of K and M as well as the position on the image surface K’ and M’. d) Calculate the ratio of the shortest distance of MK between the datum and the projection surface. e) Compute the directional derivatives for the parameter curves. Is point P (ϕ = 45◦ , λ = 50◦ ) regular? f) Find the Gaussian quantities of the datum surface. g) Deduce the square of the line element ds2 an.

Exercise 2 The formulas of the Mercator-Sanson projection are:   x = Rϕ x(φ, λ) = y  = R λ cos ϕ . Calculate for point P (ϕP = 30◦ north, λP = 45◦ east) a) the distortion in the direction of the parametric curves and b) the principal distortions. Exercise 3 Find the mapping equations (in azimuthal coordinates) of an equal area conic projection with one coinciding equidistant parallel. a) Derive the relationship between the radius and the parameter δ. b) Calculate for A (λ = 0◦ , ϕ = 40◦ and B (λ = 30◦ east, ϕ = 60◦ north) – the principal distortions, – the direction of the equidistant line element, – and the direction of the maximal distorted azimuth – as well as the maximal distortion of the angle assuming normal alignment and a pol distance of δg = 50◦ (R = 6370 km).

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c) Calculate the Euclidean distance A B  between the two points (on the image surface). Compare the shortest distance of the datum surface with the shortest distance on the image surface.

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Script ”Map Projections”

Exercise 4 a) Determine the mapping equations of an equal area conic projection in normal alignment with two equidistant parallels (Projection of Albers). b) Show the graticule for the two equidistant parallels ϕ1 = 40◦ north and ϕ2 = 70◦ north, assuming the origin of the coordinate system λKN = 0◦ . 0◦ −180◦ Interval: 30◦ , Map-scale: = 6371 km

≤ ϕ ≤ 90◦ ≤ λ ≤ 180◦ 1 : 100 000 000, Radius of the sphere: R

c) Calculate the principal distortions for the two point P1 (ϕ1 = 40◦ north, λ1 = 120◦ east) and P2 (ϕ2 = 60◦ north, λ2 = 120◦ west). Where are the loci sharing the same properties? Exercise 5 A conformal oblique conic projection with one coinciding parallel is presumed. The area between the parallels δ1 = 30◦ and δ2 = 70◦ has to be mapped in such a way, that the distortions for δ1 and for δ2 are equal. a) Search the mapping equations (in polar coordinates). Compute the polar distance of the coinciding parallel? b) Calculate the principal distortions for P (bP = 40◦ north, lP = 100◦ east) if the origin of the cartographic coordinate system KN is given by the geographic coordinates ϕKN = 50◦ north and λKN = 130◦ west. c) Calculate the intersection angle between the azimuth A = 50◦ and the vertical in P  on the image plane. Exercise 6 a) The general mapping equations for an azimuthal projection with an orthogonal parameter system has to be computed (based on a sketch). Determine the general principal distortions. b) Find the specific equations for a equal area azimuthal projection.

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c) Calculate for point P (ϕP = 40◦ north,λP = 80◦ east) – the specific principal distortions – the direction of the equidistant line element on the datum and the projection surface – the direction where the angle is not distorted (datum and image surface) – the maximal distortion of the azimuth – the angle with the maximal distortion and the amount – the intersection of the graticule on the projection surface. Geographic coordinates of the principal point H: ϕH = 30◦ north, λH = 50◦ west. Exercise 7 a) Deduce the mapping equations of an oblique conformal azimuthal projection with one coinciding point (hint: no distortion for the principal point). b) Derive the position vector x (ϕ, λ) in geographical coordinates, if the coordinates for the principal point are: ϕH = 0◦ and λH = 0◦ (transversal alignment). Hint: x tan = 2



sin x 1 − cos x = , 1 + cos x 1 + cos x



   x dx = ln tan , sin x 2

1 d tan x = dx cos2 x

Exercise 8 a) Derive the mapping equations of an azimuthal projection, projecting the sphere from its center on a coinciding plane. What is the name of this projection and what are the specific properties? b) Compute for P (ϕP = 60◦ north, λP = 60◦ east) – the principal distortion

98

Script ”Map Projections” – the distortion of the area – the maximal distortion of the azimuth presuming a principal point with the geographical coordinates ϕH = 30◦ north and λH = 40◦ east. c) Calculate the intersection angle of the graticule on the image plane. d) Compute the azimuth (datum surface) of the maximal distorted bearing?

Exercise 9 Deduce the properties (equal area, conformal, equidistant, . . . ) of the below depicted azimuthal projections: H

P’

H

P’

P

P

M

M

(a) Orthographic projection H

(b) Stereographic projection H

P’

P

P

M

(c) Projection of Lambert

Exercise 10

P’

M

(d) Equidistant azimuthal projection

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99

a) Derive the mapping equations of an equal area cylinder projection with two coinciding parallels. b) Determine the equidistant parallel ϕl in such a way that the below equation is fulfilled: 2 ' (an − 1)2 → min. n=1

Mapped area:

◦

−20 ≤ u ≤ 20◦

c) Calculate the rectangular coordinates x and y  on the image surface of P (ϕP = 20◦ north, λP = 80◦ west). – The origin of the image surface KN has the geographic coordinates ϕKN = 20◦ north and λKN = 60◦ east. – The parallels u1 = 20◦ and u2 = −20◦ are mapped equidistant. d)

– Determine the mapping equations in geographic coordinates for normal attitude. – Which areas are depicted advantageous or disadvantageous respectively? – Which size has the map representing the whole world (scale 1 : 150 000 000)?

Exercise 11 Based on a map which is the result of a conformal projection of the ellipsoid of revolution to the plane a captain derives the course angle of his passage from San Juan (Puerto Rico) (ϕJ = 18.28◦ , λJ = 66.07◦) to Lisbon (ϕL = 38.43◦ , λL = 9.08◦ ). The course angle shall be the same for the whole journey. a) Calculate the difference of the course angle, if the map is not derived from a ellipsoid of revolution but from a sphere (radius = 6 370 km) b) How is the difference of the length of the loxodrome between the ellipsoid of revolution and the sphere? What is the length of the loxodrome if we calculate with the course angle of the sphere? c) And finally, what is the difference in the distance between the loxodrome and the shortest distance on the sphere?

100

Script ”Map Projections”

General information: a = 6377.4 km; e2 = 0.0066744 

   x π dx , = ln tan + cos x 2 4



   e  π Φ 1 − e sin Φ 2 M(Φ) · dΦ = ln tan + N(Φ) cos Φ 4 2 1 + e sin Φ

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101

Exercise 12 Deduce: The product of the distortion of the azimuth and the square of the scale distortion is constant and equals the product of the principal distortions and thus the distortion of area. dβ  2 λ =V dβ Hint: Use the following equations: sin β  =

b sin β λ

cos β  =

a cos β λ

Exercise 13 Given is a map from South America in a transversal cylinder projection with one coinciding equidistant parallel (see supplemantal sheet). Further on the projection comprises the below properties: • The cylinder coincides with the sphere at λ = 60◦ west. • The origin is defined by the intersection of the coinciding curve with the equator. • The geographical latitude of the principal point is λ = 30◦ east. • The mapping equations relating to the origin (αKN ) are:   x = R(αKN − α) x(α, δ) = y  = R( π2 − δ) (R = 6371 km) a) Outline (in correct position) the polar triangle of point P (marked in the map). b) Find the coordinates on the projection surface of P . c) Calculate the below distoritons: • Principal distortions (elements of the distortion ellipse) • Distortion of the area

102

Script ”Map Projections” • The maximal distorted azimuth and the amount (on the datum and the image surface) • The angle with the maximum distortion and the amount of the distortion.

d) Compute the intersection angle of the images of the (geographic) graticule in P . e) What is the azimuth of the maximal distorted bearing in P ? f) Depict the distortion ellipses for the origin of the coordinate system and for point P . Exercise 14 has been omitted Exercise 15 Given is the trapezoidal projection of Eckert (known as Eckert II; see page 126):     2π  x 2 − = sign(φ) 4 − 3 sin |φ| 3 . (A.4) x (φ, λ) = λ 4 − 3 sin |φ| y  = √26π • Outline the meridian for λ = ±45◦ . (Hint: The slope of the image of the meridian is constant within the interval φ ∈ [0, π/2] and φ ∈ [0, −π/2] respectively.) • Show: It is an equal area projection.

Appendix B Mathematical explanations B.1

Trigonometry

Relationships for the sum and difference of arguments sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y ∓ sin x sin y tan x ± tan y tan(x ± y) = 1 ∓ tan x tan y

(B.1) (B.2) (B.3)

Relationships for multiple arguments sin 2x = 2 sin x cos x = cos 2x = cos2 x − sin2 x Further general relationships     π π sin − x = cos +x ; 4 4   π sin + x = cos x ; 2   π − x = cos x ; sin 2

2 tan x 1 + tan2 x

   π π cos − x = sin +x 4 4   π cos + x = − sin x 2   π cos − x = sin x 2

103

(B.4) (B.5)



(B.6)

(B.7) (B.8)

104

Script ”Map Projections”

Figure B.1: sin(x) and cos(x) Relationships for the half of the argument  x 1 − cos x = ± sin 2 2  x 1 + cos x = ± cos 2 2  x 1 − cos x sin x 1 − cos x tan = ± = = 2 1 + cos x 1 + cos x sin x

B.2

(B.9) (B.10) (B.11)

Spherical trigonometry

Spherical law of cosines : cos c = cos a cos b + sin a sin b cos (γB.12) sin b sin c sin a = = (B.13) Spherical law of sines : sin α sin β sin γ Spherical law of cotangents : cos b cos γ = sin b cot a − sin γ cot(B.14) α

B.3

Integrals and series 3

(1 − x)− 2 = 1 +

3·5 2 3·5·7 3 3 x+ x + x . . . (|x| < 1) 2 2·4 2·4·6 sin(x) = x −

x3 x5 + −... 3! 5!

(B.15) (B.16)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch x2 x4 x6 + − + ... 2! 4! 6! x2 x3 x + + ... ex = 1 + + 1! 2! 3!  1 1 sin2 ax = x − sin 2ax 2 4a  1 1 sin 2ax cos2 ax = x + 2 4a  1 1 3 sin 2ax + sin 4ax sin4 ax = x − 8 4a 32a cos(x) = 1 −

105

(B.17) (B.18) (B.19) (B.20) (B.21)

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Script ”Map Projections”

Figure B.2: The concept of shape (Boyce & Clark)

Appendix C Formulary C.1

Scale distortion v + dv = const.

v = const.

v = const.

v + dv = const.

The infinitesimal line element

β

ds

’

E’ du

u + du = const.

ds

E du

u + du = const.

β’ u = const.

u = const. G’dv

G dv

(a) datum surface

Definition: m=

(b) image surface

line element on datum surface ds = ds line element on image surface

or m2 =

ds2 E  du2 + 2 F dudv + G dv 2 = ds2 Edu2 + 2 F dudv + Gdv 2 107

108

Script ”Map Projections”

Distortion along the parameter curves u-Linie: v = const. ⇒ dv = 0  m2u = EE = h2 v-Linie: u = const. ⇒ du = 0  m2v = GG = k 2 Scale distortion and azimuth " m2 = E 

du ds

#2

" #2 dv du dv + 2F  + G ds ds ds

!        − F E − 2F F + GE E EF EG E cos2 β+2 √ sin2 β ⇒ m2 = sin β cos β+ − E EG − F 2 E E EG − F 2 ⇒ m2 = A cos2 β + 2B sin β cos β + C sin2 β mit A =

E , E

B=

EF  − F E  √ E EG − F 2

and C =

EG − 2F F  + GE  E  − EG − F 2 E

i. e. the units A, B and C depend on E, F and G. ⇒ m = m(u, v, β) Directions with extreme scale distortion (principal distortion) Datum surface:

m = m(u, v, β) Condition: ⇒ tan 2βh =

dm dβ

=0

√ 2(EF  −F E  ) EG−F 2 2F (EF  −F E  )+E(E  −EG )

⇒ tan 2βh2 = tan(2βh1 ± π) ⇒ 2βh2 = 2βh1 ± π ⇒ βh2 = βh1 ± Image surface:

m = m(u, v, β )

π 2

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch Condition: ⇒ tan 2βh =

dm dβ 

109

=0

√ 2(F E  −EF  ) E  G −F 2 2F  (F E  −EF  )+E  (EG −GE  )

⇒ βh 2 = βh 1 ±

π 2

Theorem 1 The principal directions are orthogonal on the datum surface as well as on the image surface. In general these are the only according orthogonal directions of the projection. Assumption: F ≡ F  ≡ 0 ⇒ tan 2βh = 0 → βh1 = 0◦

and βh2 = 90◦

⇒ tan 2βh = 0 → βh 1 = 0◦

and βh 2 = 90◦

Theorem 2 Are F ≡ 0 and F  ≡ 0, then the parameter curves are coincident with the directions of the principal directions. Amount of extreme scale distortion (principal directions a and b) Assumption: F ≡ F  ≡ 0 Scale distortion along the parameter curves:  u-Linie (v = const.): a = h = v-Linie (u = const.): b = k = Scale distortion in an arbitrary direction: m2 = a2 · cos2 β + b2 · sin2 β

C.2

Area distortion

The infinitesimal area



E E

G G

v = const.

v = const.

v + dv = const.

Script ”Map Projections” v + dv = const.

110

u + du = const. E’ du

E du

u + du = const.

dO

τ

dO’

τ’ u = const.

u = const.

G dv

G’dv

(a) Datum surface

(b) Image surface

Definition: V =

infinitesimal area on image surface dO  = dO infinitesimal area on datum surface

Area distortion as a function of h and k (or mu and mv ): V =h·k·

sin τ  sin τ

Area distortion as a function of a and b: V =a·b

Theorem 3 The area distortion equals the product of the principal directions.

C.3

Distortion of angles

Precondition: F ≡ F  ≡ 0

β’

dβ

F 0 P(u,v)

u = const. 2. HVR

G dv

dβ ’

F’ 0 P’(u,v)

(a) im Original

C.3.1

ds ’

β

ds

E du

E’ du

v = const. 1. HVR

111

v = const. 1. HVR

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

u = const. 2. HVR

G’dv

(b) im Bild

Distortion of the angle element

Definition: Infinitesimal angular distortion =  tan β  = ab tan β ⇒ dβ · m2 = a · b dβ

dβ  dβ

Theorem 4 The product of the distortion of the azimuth and the square of the scale distortion is constant and equals the product of the principal distortions and thus the distortion of area. Direction of extreme distortion of the angle     Minimum Maximum dβ  2 = ⇐⇒ m = Maximum dβ Minimum F ≡ F  ≡ 0 ⇒ direction of extreme scale distortion = parameter curves Theorem 5 The directions of extreme distortion of the angle coincide with the principal direction. Amount of extreme distortion of the angle   u-Linie: m = a → dβ =  dβ u−Linie v-Linie: m = b → dβ = dβ v−Linie

b a

< 1 f¨ ur a > b

a b

>1

112

Script ”Map Projections” dβ  dβ

=1⇒ datum surface : image surface : conclusion :

u−Linie

d β’ =1 dβ −βm

m2 tan2 βm  tan2 βm  βm + βm

= a·b = ab = ab = 90◦

d β’ 1 dβ Figure C.1: The correlation of scale distortion and direction assuming F ≡ F  ≡ 0 und a > b

C.3.2

Distortion of the azimuth

Definition: ω = β − β Maximum distortion of the azimuth: Condition:

dβ  dω =0⇒ =1 dβ dβ

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Theorem 6 The distortion of the azimuth has its maximum where the angle element is not distorted. Maximum distortion of the azimuth:  ωm = βm − βm

oder

sin ωm =

b−a b+a

Maximal distorted angle (2 · |βm |): The amount of the maximal distorted angle: Ωm = 2 · ωm

u−Linie −βm βm v−Linie P(u,v)

Appendix D Additional explanations D.1

The Sphere

Figure D.1: The sphere (Gaussian parameters)

114

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D.2

115

Alignment of the surfaces

Figure D.2: Alignment of the surfaces (cone, cylinder, plane → normal, transversal, oblique)

116

D.3

Script ”Map Projections”

Projection on a coinciding plane

Figure D.3: Sphere and coinciding plane surface

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D.4

Projection on a cone surface

Figure D.4: Sphere and cone

117

118

Script ”Map Projections”

Figure D.5: Cone surface and plane

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D.5

Projection on a cylinder surface

Figure D.6: Sphere and coinciding cylinder

119

120

D.6

Script ”Map Projections”

Tissot’s ellipse

Figure D.7: Tissot’s ellipse

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D.7

Specific projections

Figure D.8: Aitoff

121

122

Script ”Map Projections”

Figure D.9: Aitoff (Indicatrix)

Figure D.10: Hammer-Aitoff

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

Figure D.11: Kavraiskii VII

Figure D.12: Mollweide

123

124

Script ”Map Projections”

Figure D.13: Mollweide (Indicatrix)

80

40 80

60 40

20

10

80

60

5 60 20

40

10

20

20

20

20 10

80

5

40

20

60

10

40

20

60

80 40 80

60

Figure D.14: Mollweide (angular distortion)

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Figure D.15: Winkel Tripel

Figure D.16: Ginsburg V

125

126

Script ”Map Projections”

Figure D.17: Miller I

Figure D.18: Eckerts projection (known as Eckert II)

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

Figure D.19: Eckert IV

Figure D.20: Mercator-Sanson projection

127

128

Script ”Map Projections”

Figure D.21: Conic projection (two standard parallels)

Index Airy, 78 Airy-Kavrajski, 78 Albers’ projection, 96 area distortion definition, 17 area distortion, 16, 17, 110 area element, 109 azimuth distortion, 112 surface curve, 8 azimuthal projection, 49 conformal, 51 equal area, 50 gnomonic projection, 52 Lambert, 50 orthographic projection, 52 stereographic projection, 51 base direction principal direction, 14 Bonne, Rigobert (1727 - 1795), 58 Cartesian coordinates azimuthal, 50 central point definition, 31 Chamberlin, Wellman, 76 conformal criterion, 48 conic projection

Albers eqal area, 43 basics, 40 conformal, 45 one standard parallel, 46 two standard parallels, 47 equal area, 41, 54 principal distortions, 41 pseudoconic, 54 coordinates azimuthal, 30 cross product, 2, 7, 9 cylinder projection scale factor, 47 cylindrical projection, 47 conformal, 48 equal area, 48 derivative directional, 7 differential equation, 50 disstortion ellipse, 21 distortion angle, 110, 111 extreme, 111 area, 16, 109, 111 azimuth, 18, 19, 98, 108 direction, 112 infinitesimal angle, 20 infinitesimal parallelogram, 17 infinitesimal triangle, 17 maximum, 14 129

130 minimum, 14 of the area, 98 scale, 11, 111 dot product, see inner product Eckert projection, 65 elliptic integral, 34 equal area criterion, 48 cylinder, 65 equidistant, 24

Script ”Map Projections” Lagrange, Joseph Louis (1736 - 1813), 9 law of cosine spherical, 6 line element, 94, 107 line scale, 86 loxodrome, 31 direction, 32 length, 32 medial Distortion, 16 Mercator projection, 89 Mercator-Sanson projection, 62

Fiorini, 78 first Gaussian fundamental quantities see Gaussian quantities, 3 Newton method, 64 First Gaussian quantities, see Gaus- Newton-Raphson method, see Newsian quantities ton method normal vector Gaussian quantities, 3, 94 see surface normal, 2 Gaussian quantity sphere, 7 orthodrome, 38 azimuth, 38 inner product, 7, 8 computation of support points, 39 integral length, 38 elliptic, 34 orthogonal parameter system, 40 specific, 104 intersection parallelogram angle infinitesimal, 9 parameter curve, 8 point surface curve, 8 regular, 3 porjection Jacobian determinant, 3 conic Jacobian matrix, 3 equal area, 96 Jordan, 78 position vector Jordan-Kavrajski, 78 cylinder coordinates, 47 principal Klingatsch, 78 distortion, 26 principal direction, 14, 109 L¨obell, 78 principal distortion, 97 Lagrange’s formula, 9

KIT, Geodetic Institute, Dr.-Ing. N. R¨osch

131

projection pseudoazimuthal, 59 azimuthal, 26 conformal, 98 equal area, 98 equidistant, 26, 98 geometric interpretation, 98 Bonne, 58 conformal, 45 conic, 94 equal area, 94 oblique, 96 one parallel, 96 two parallels, 96 cylinder, 93 equal area, 65 exercise, 4 generalized, 62, 65 cylindrical conformal, 48 equal area, 27, 41, 61 equidistant, 24, 26 equator, 65 meridians, 65 generalized, 54 Mercator, 89 Mollweide, 62 oblique, 29 orthogonal parameter system, 40 pseudocylindrical, 59 equal area and elliptical meridians, 62 equal area and equidistant parallels, 61 Stab-Werner, 59 projection of Apian, 65

azimuthal conformal, 51 equal area, 51 conformal azimuthal proj., 51 gnomonic projection, 52 Robinson projection, 70 Robinson, Arthur H. (1915–2004), 70 Romberg’s method, 34

radius

vector product, see coss product7

scalar product, see inner product scale distortion, 11, 111 along the parameter curves, 109 arbitrary direction, 109 as as function of the azimuth, 13 azimuth, 108 extremes, 109 parameter curve, 108 series expansion, 36, 104 spherical trigonometry, 104 Stab-Werner-projection, 59 Stereographic projection, 90 surface analytical, 3 surface normal, 2, 7 Tissot’s Indikatrix, 21 Tissot, Nicolas Auguste (1824 - 1890), 21 triangle infinitesimal, 9 trigonometric funcions half argument, 104 trigonometric functions for sum and difference, 103 general relationships, 103 multiple arguments, 103 trimetric projection, 76

132

Script ”Map Projections” see cross product, 2