1. Dr. A‘kif Al_Fugara Alalbayt University
Principles of Geodesy
Basic Spherical Geometry- Part I
Spherical Triangle
Dr. A‘kif Al_Fugara
Alalbayt University

2. Basic Spherical Geometry
Sphere is:
A set of points in three-dimensional space equidistant from a point called
Center of the sphere.
Radius: The distance from the center to the points on the sphere.
Notice that we are talking about the surface of a ball, and not the ball itself.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

3. Basic Spherical Geometry
Lines and spheres:
The line and the sphere can miss each other
The line can intersect the sphere in only one point:
In that case the line is tangent to the sphere.
The line pierces the sphere in precisely two points:
In particular a line cannot lie in the sphere.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

4. Basic Spherical Geometry
Antipodal points:
Intersection of the sphere with line passing through the center of the sphere.
The best-known example of antipodal points is the North and South poles on the Earth.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

5. Definitions
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

6. Basic Spherical Geometry
Planes, spheres, circles, and great circles
In the uninteresting case the plane and the sphere miss each other.
The plane and sphere meet in a single point.
In this case the plane is tangent to the sphere at the point of intersection.
The sphere and the plane meet in a circle.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

7. Basic Spherical Geometry
Great circles:
Intersection of the sphere with plane passing through the center
A geographic example of a great circle is the equator.
The meridians of longitude form exactly half a great circles
The parallels of latitude are small circles, except for the equator
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

8. Basic Spherical Geometry
Shortest distance between two points on the sphere:
along the segment of the great circle joining them
On any surface the curves that minimize the distance between points are called geodesics.
lines are the geodesics on the plane,
great circles - on the sphere.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

9. Basic Spherical Geometry
Incidence Relations on a Sphere:
A and B are not antipodal points. Then A, B, and C do not lie on the same line in space, and consequently determine a unique plane.
the intersection of the plane with the sphere is a great circle containing A and B.
Thus A and B determine a unique great circle.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

10. Basic Spherical Geometry
Incidence Relations on a Sphere:
If A and B are antipodal points, then A, B, and C lie on the same line in space.
Any plane, which contains this line, determines a great circle that must contain A and B.
Thus there are infinitely many great circles containing A and B if they are antipodal.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

11. Basic Spherical Geometry
Incidence Relations on a Sphere:
Now suppose we have two great circles.
Each of these is the intersection of the sphere with a plane through the center.
These planes must intersect in a line in space, which of course intersects the sphere in two antipodal points. Thus:
Two distinct great circles meet in exactly two antipodal points.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

12. Basic Spherical Geometry
Spherical distance:
A and B are two points on the sphere
The distance between them is the distance along the great circle connecting them.
The circle lies totally in a plane
If the angle ACB is a, and if a is measured in radians, then the distance between A and B is given by
d(A,B) = R a
where R is the radius of the sphere.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

13. Basic Spherical Geometry
Isometry of the sphere:
Mapping of the sphere to itself that preserves the distance between points.
Examples:
rotation of the sphere around one of the sphere's diameters.
antipodal map, which maps a point onto the point on the other side of the sphere- the antipodal one.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

14. Basic Spherical Geometry-Part II
Geodesy
Basic Spherical Geometry-Part II
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

15. Basic Spherical Geometry
Lunes or biangles :
Simplest figure on a sphere:
The two sides are halves great circles
The vertices of a lune are antipodal points.
The two angles of a lune are equal.
Why is it called a lune?
The name comes from the Latin word luna, which means moon.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

16. Basic Spherical Geometry
Angles on the sphere:
The angle between two curves is the angle between the tangent lines to the curves at the point of intersection
Usually measured in radians.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

17. Basic Spherical Geometry
In the case of a lune:
The angle between the great circles at either of the vertices is simply the angle between the planes that define the great circles
it does not matter at which vertex the measurement is made.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

18. Basic Spherical Geometry
Part of a sphere
Lunar angle 
Area of the lune
Sphere
360=2 radians
4R2
Hemisphere
180=  rad
2R2
1/4 sphere
90 = /2 rad
R2
1/8 sphere
45 = /4 rad
R2/2
1/n sphere
360/n=2/n rad
4R2 /n =2 R2
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

19. Basic Spherical Geometry
Spherical triangles:
consists of three points called vertices,
the arcs of great circles that join the vertices- sides, and
the area that is enclosed therein.
Ambiguity
There are two segments of the great circle that join each pair of points.
The resulting figure is the boundary of two different regions
the outside (green)
the inside (brown)
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

20. Basic Spherical Geometry
Small triangles:
No pair of the three vertices are antipodal
The sides are short segments of great circles that join the vertices
There is one hemisphere that contains the three points of the triangle
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

21. Basic Spherical Geometry
Area S of a spherical triangle:
The vertices of the triangle form 3 lunes:
AA’ with angle 
BB’ with angle 
CC’ with angle 
Area of each lune:
Area AA’=2R2
Area BB’=2R2
Area CC’=2R2
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

22. Basic Spherical Geometry
Sum of the areas of the three lunes:
Area of the spherical triangle:
Spherical excess:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

23. Basic Spherical Geometry
Sine and Cosine laws for spherical triangles:
The surface angles of the spherical triangle are denoted by capital letters
The sides of the triangle:
Denoted by small letters
Can be measured in linear units or,
By their angles subtended at the center of the sphere
Each side is labeled by the same letter as the opposite surface angle (only different case)
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

24. Basic Spherical Geometry
Law of Sines
Law of Cosines
Sin—Cos Relation:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

25. Sine rule
The Sine rule can be used to find an angle if two sides and an angle are known, or to find a side if two angles and a side are known.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

26. Cosine rule
The Cosine rule allows the length of one of the arcs of a spherical triangle to be evaluated if the other two arcs and the angle opposite the arc are known.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

27. Sine & Cosine rule
The Sine and Cosine rule allows the length of one of the arcs of a spherical triangle to be determined if the other two arcs and the two angles are known.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

28. Fundamentals of Geodesy
Spherical Earth
Dr. A’kif Al_Fugara, Dept. of Surveying Eng., AABU

29. Spherical Earth Definitions:
Meridian: Great circle passing trough the North and South poles
Equator: Great circle normal to the rotation axis of the Earth.
Parallels: Small circles parallel to the equator.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

30. Spherical Earth Geographic coordinates: Latitude  Longitude 
from 0 to  90  (North or South of Equator)
Longitude 
from 0 to  180  (East or West of Greenwich Meridian)
Height H
Above MSL or a reference datum
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

31. Spherical Earth Definitions:
Distance between two points on the sphere is the length of the arc of great circle connecting the points.
Azimuth:
Angle measured clockwise from North to the line.
Angle between the meridian and the arc of great circle that represents the distance; measured clockwise from the meridian
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

32. Spherical Earth Consider the triangle (A,B,North Pole)
Sides of the triangle:
SAB
90-A
90-B
Angles in the triangle:
Azimuth 
Difference in longitudes 
The sine and cosine laws for spherical triangles can be applied for coordinate computations
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

33. Spherical Earth Direct problem: Given:
Geographic coordinates of the first point A A, A
Azimuth toward second point B AB
Spherical distance SAB
Compute the coordinates of the second point B B and B
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

34. Spherical Earth Solution- Direct problem:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

35. Spherical Earth Inverse Problem: Given: Compute:
Geographic coordinates of point A: A, A
Geographic coordinates of point B: B, B
Compute:
Azimuth of the line AB AB
Spherical distance SAB
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

36. Spherical Earth Solution- Inverse problem:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

37. Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

38. Spherical excess
A spherical triangle, differs from a plane triangle in that the sum of the angles is more than 180o.
Where:
T spherical triangle area
R radius of sphere
 spherical excess
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

39. Spherical excess
A spherical triangle area, that its one sides not exceed about 60 km, can be calculated by considering it equal to area of plane triangle, namely:
If spherical triangle has sides longer than 60 km, the radius of sphere should be considered:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

40. Differential rectangle
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

41. Differential rectangle
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

42. Differential rectangle
Surface area of a spherical part that enclosed between two parallels (1, 2) and two meridians (1, 2) is given as follows:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

43. Exercise
A petroleum company has won the right to explore for oil in a region has the following data:
1 = 34° 30 N, 2 = 35° 10 N
1 = 38° 50 E, 2 = 39° 35 E
Compute the area of determined region by considering the Earth’s shape is a sphere.
R = 6371 km
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

44. Solution
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

45. Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

46. Exercise
Finding the Qibla direction (direction to Mecca) from Jakarta (Indonesia).
Where:
Mecca (21.26°N, 39.49°W)
Jakarta (6.08°S, °E)
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

47. Solution Using the diagram below, set C to Jakarta and B to Mecca. A B
Greenwich
Equator
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

48. Solution
We are looking for angle (C). This is the angle between the line of longitude (b) which faces due north and (a), the arc of the great circle joining point C to point B.
The arcs of the great circles (c and b) are known. They are simply the polar distances of points B and C respectively (90° minus the Latitudes of these points).
Angle (A) is also known: it is the difference in longitudes between points C and B. The only way that the Sine rule can be used is as follows:
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

49. Solution
(C) is the angle required. It will be west of North (angle ACB on the diagram).
(A) is the longitude difference between Mecca and Jakarta. In this case the value is 66.96°.
(c) is the polar distance of Mecca (90° ° = 68.74°).
(a) is the distance between Mecca and Jakarta. This is found by using the Cosine rule and has a value of ° (corresponding to a distance of 7912 km).
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

50. Solution
To face Mecca from Jakarta, one must face North and turn anti-clockwise through an angle of 65.04°.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

51. Exercise Find the distance between London (UK) and Baghdad (Iraq).
Where:
London (51.30°N, 0.10°W)
Baghdad (33.20°N, 44.26°E)
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

52. Solution
If we use the diagram below then B is London while C is Baghdad. A S C B
Greenwich
Equator
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

53. Solution
We seek (a), the angular distance along a great circle between London and Baghdad. This is the shortest distance between these two cities along the surface of the Earth. It is shown in red in the diagram.
A (the difference in Longitude between London and Baghdad) is 44.26°E °W = 44.36°. The two cities lie on different sides of the Greenwich meridian (just) so we add the longitudes.
(b) is the polar distance of Baghdad given by 90° minus the latitude of Baghdad (90° ° = 56.80°).
(c) is the polar distance of London: 90° minus the latitude of London (90° ° = 38.70°).
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

54. Solution
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

55. Exercise
A submarine cable will be extended between two cities I and J that located on both sides of a sea and have the following geodetic coordinates:
I (34° 55 N, 56° 10 E)
J (33° 56 N, 130° 48 E)
Calculate:
The minimum cable length in kilometers.
The direction or azimuth that should be extended this cable from point I.
R = 3671 km.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

56. Submarine cable
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

57. Right spherical triangle
Omitting the right angle, we write the remaining five components in order on a pie diagram, but replace the three components not adjacent to the right angle by their complements.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

58. Right spherical triangle
Napier’s rule:
The following relationships hold for right spherical triangle between the five parameters in the circle.
The sine of any component in the circle is equal to the product of either
(a) the tangent of adjacent component or
(b) the cosines of the opposite component
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

59. Exercise
A ship is sailing along a great circle from point A on the equator at an angle equal to 60°.
Calculate:
The longitude difference between point A (departure) and B (destination) that located on the path of the ship and has the latitude 30° N.
The area enclosed by ship’s path, equator and meridian of point B.
Consider the radius of sphere is R = 3671 km.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

60. Solution Spherical excess equals:  = 5o 15 51.8
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU

61. Legendre’s theorem
If each of the angles of a spherical triangle whose sides are smaller than 60 km be diminished by one third of the spherical excess, the triangle may be solved as a plane triangle whose sides are equal to the sides of the spherical triangle, and whose angles are these reduced angles.
Dr.A'kif Al_Fugara, Dept. of Surveying, AABU