1. 16. Latitude and Longitude
A little bit of spherical trigonometry: great circles, the Haversine Law, Rhumb lines and Mercator maps
Contest Algorithms: 16. Lat & Long

2. Latitude and Longitude

3. Contest Algorithms:16. Lat & Long

5. Great Circles
The largest circle that can be drawn on the surface of the earth.
Equator
Meridian/ Longitude line
Another Great Circle

6. Small Circles Parallel/ Latitude line Another Small Circle
All latitude lines (parallels) are small circles except for the equator

7. Nautical Miles
A nautical mile (NM) is the distance travelled on a great circle when moving through 1 minute of arc.

8. Length of the Equator
There are 360o in a circle and 60' (minutes) in a degree.
No. of minutes in a circle: 360 x 60 = 21600'
NM length of the equator: NM
1 NM == km (by int. agreement)
km length of the equator = * = 40,003 km
actual is 40,075 km
Contest Algorithms:16. Lat & Long

9. From Angle to NM
Calculate the shortest distance from 60o W to 10o E along the equator.
Smallest angle along the equator between the two positions is =70o
Shortest distance
= 70 x 60
= 4200 NM
convert degrees to
minutes

10. Latitude Distance
Equator
Latitude 41o N N S
41o R r
As you travel around a line of latitude (small circle) the distance travelled is shorter than the distance covered on the equator (great circle).
Radius of the small circle: r
Radius of the Equator: R
Latitude angle: 
Then r = R cos 
So the small circle is smaller than the great circle by a factor of cos . N r R

11. Calculate a Latitude Distance
Calculate the distance along the line of latitude from (41o N, 36o W) to (41o N, 155o E) .
Smallest angle is not 191o, but 169o
Converting this angle into minutes:
= 169 x 60 = 10140'.
Since the measurement is along a small circle the distance is reduced by a factor of cos 41o.
Distance in nautical miles
= x cos 41o = 7653 NM N
36o W
155o E 0o

12. Review of Plane Trigonometry
Cosine Law: A c 2 = a + b - ab
cos C c b
Sine Law: a
sin A = b B c C B C a
Angles A, B and C; Sides a, b and c

13. Spherical Trigonometry
Great circles arcs / sides / angles: a, b, c
this assumes a unit sphere; otherwise multiple by radius R to get lengths
0 < a + b + c < 3π
Face/surface angles / bearings: A, B, C
π < A + B + C < 3π
Contest Algorithms:16. Lat & Long

14. Spherical Trig. Laws Sine Law sin a A = b B c C Cosine Law for sides
a b c are sides
A B C are angles
Cosine Law for sides
Cosine Law for angles
cos a = b c +
sin A B C
cos A = - B C +
sin a b c

15. Calculating Distance P
Calculate the air distance from New York to Moscow
We know their coordinates.
New York: Latitude = 40o47'N, Longitude = 73o58'W
Moscow: Latitude = 55o45'N, Longitude = 37o42'E
The air distance = great circle arc from N to M == p m n N M p
Contest Algorithms:16. Lat & Long

16. Views of the Triangle P P P n m M N m n M N View from above earth's
center n P m M
73o58'W
37o42'E N
New York's
longitude
Moscow's
longitude m n
55o45'N
40o47'N N M
New York's
latitude
Moscow's
latitude
prime
meridian
equator
earth's
center
earth's
center
Contest Algorithms:16. Lat & Long

17. Calculate cos p = cos m cos n + sin m sin n cos P Sides m and n:
m = 90o – lat of New York = 49o13' =
cos m = ; sin m =
n = 90o - lat of Moscow = 34o15' =
cos n = ; sin n =
Face/surface angle P
P = long NY + long MO = 73o58'W + 37o42'E = 111o40' =
cos P =
Contest Algorithms:16. Lat & Long

18. cos p = cos m cos n + sin m sin n cos P Substitute:=
p = cos = o == *60 minutes = 405.3'
= NM == * = km
Google reports:
Contest Algorithms:16. Lat & Long

19. The Haversine Law cos a = cos b cos c + sin b sin c cos A
Inaccurate when a is small
Instead substitute:
cos θ = 1 − 2 hav θ
cos(a − b) = cos a cos b + sin a sin b
Produces:
hav a = hav(c-b) + sin c sin b hav A
Contest Algorithms:16. Lat & Long

20. The Haversine Formula
Considers the special case when A is the north pole, and B and C are the two points whose separation d is to be determined.
In that case, c and b are 90° − their latitudes, A is the longitude separation Δlon, and a is the desired d/R.
Note that sin(π/2 − φ) = cos(φ)
Contest Algorithms:16. Lat & Long

21. hav(a/R) = hav(lat2 - lat1) + cos(lat1) cos(lat2) hav(lon2 – lon1)
lati and lon i are the lattitide of the two points
R is the Earth's radius
hav(angle) = sin2(angle/2) [ = (1 – cos(angle))/2 ]
Let rhs of 1st equation be h, then
hav(a/R) =h, so sin2(a/2R)= h
a = 2R arcsin( √h)
Contest Algorithms:16. Lat & Long

22. Code see LatLongTests.java
public static double haversine(double lat1, double lon1, double lat2, double lon2) { double dLat = Math.toRadians(lat2 - lat1); double dLon = Math.toRadians(lon2 - lon1); lat1 = Math.toRadians(lat1); lat2 = Math.toRadians(lat2); double a = Math.pow(Math.sin(dLat/2), 2) + Math.pow(Math.sin(dLon/2), 2) * Math.cos(lat1) * Math.cos(lat2); double c = 2 * Math.asin(Math.sqrt(a)); return EARTH_RADIUS_KM * c; } // end of haversine()
Contest Algorithms:16. Lat & Long

23. public static double fromDMS(String dmsStr) /
public static double fromDMS(String dmsStr) /* Obtain an angle from a degrees, minute and secs character string. see e.g S W */ { if (dmsStr == null) { System.out.println("String is null"); return 0; } String[] dms = dmsStr.split("\\s+"); if (dms.length != 4) { System.out.println("Incorrect DMS format in \"" + dmsStr +"\""); :
Contest Algorithms:16. Lat & Long

24. int degrees = parseInt(dms[0]); if (degrees < 0) { sign = -1;
int sign = 1;
int degrees = parseInt(dms[0]);
if (degrees < 0) {
sign = -1;
degrees *= -1; }
int mins = parseInt(dms[1]);
double secs = parseInt(dms[2]);
char suffix = dms[3].toUpperCase().charAt(0);
if (suffix == 'S' || suffix == 'W')
sign *= -1;
if (mins < 0 || mins >= 60) {
System.out.println("Minutes out of range; using 0");
mins = 0;
if (secs < 0 || secs >= 60) {
System.out.println("Seconds out of range; using 0");
secs = 0;
return ( sign*(Math.abs(degrees) + mins/60d + secs/3600d));
} // end of fromDMS()
Contest Algorithms:16. Lat & Long

25. Usage > java LatLongTests Great circle distance: 7507.8
public static void main(String[] args) { // New York coordinates (from double nyLat = fromDMS(" N"); double nyLong = fromDMS(" W"); System.out.printf("New York lat,long: %.3f, %.3f\n", nyLat, nyLong); // , // Moscow coordinates double mLat = fromDMS(" N"); double mLong = fromDMS(" E"); System.out.printf("Moscow lat,long: %.3f, %.3f\n", mLat, + mLong); // , System.out.printf("Great circle distance: %.1f\n", haversine(nyLat, nyLong, mLat, mLong) ); : // more examples }
Contest Algorithms:16. Lat & Long

26. Distance from NY to Moscow (2)
Coordinates:
New York: Latitude = 40o47'N, Longitude = 73o58'W
Moscow: Latitude = 55o45'N, Longitude = 37o42'E
Calculation:
diffLat = =
diffLon = =
h = sin²(diffLat/2) + cos(lat1) * cos(lat2) * sin²(diffLon/2); // in radians
h = ( * * ) = =
a = 2*6371*arcsin(√h)
a = * = , which agrees with Google's result
Contest Algorithms:16. Lat & Long

27. Navigational Terminology
Azimuth / bearing / true course: the angle a line makes with a meridian, taken clockwise from north
North=0°, East=90°, South=180°, West=270°
the A, B, C angles
Geodesic: The smaller great circle arc through two given points
the shortest distance between the two points
the a, b, c "lines”

28. Bearing from NY to Moscow
Use the sine law:
sin 𝐴 sin 𝑎 = sin 𝐵 sin 𝑏 = sin 𝐶 sin 𝑐
Bearing for New York (N) in terms of the polar data:
sin 𝑁= sin 𝑃 sin 𝑝 ∙ sin 𝑛
N = sin-1( sin( ) / sin(67.51) * )
= sin-1(0.9293/ * ) = sin-1(0.5660)
= 34.48o
Contest Algorithms:16. Lat & Long

29. Confirmation using earthdirections.org
Contest Algorithms:16. Lat & Long

30. Code
see LatLongTests.java
public static double bearing(double lat1, double lon1, double lat2, double lon2) { double dLon = Math.toRadians(lon2 - lon1); lat1 = Math.toRadians(lat1); lat2 = Math.toRadians(lat2); double y = Math.sin(dLon) * Math.cos(lat2); double x = (Math.cos(lat1) * Math.sin(lat2)) - (Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon)); return (Math.toDegrees(Math.atan2(y, x))+360)%360; } // end of bearing()
Contest Algorithms:16. Lat & Long

31. Usage
// in main() System.out.printf("Bearing from NYC to Moscow: %.3f degrees\n", bearing(nyLat, nyLong, mLat, mLong) );
> java LatLongTests
Great circle distance:
Bearing from NYC to Moscow: degrees
Contest Algorithms:16. Lat & Long

32. The Rhumb Line
The two points are located on a Mercator map, and a straight line (the rhumb line) is drawn between them.
Contest Algorithms:16. Lat & Long

33. Mercator vs. From Space Mercator map View from Space meridians
latitude
lines
This is the Great Circle Track that Charles Lindbergh flew
Owl – Shortest distance between two points is a curve (great circle)
View from Space

34. The Mercator Projection
Let the globe be a spherical balloon that is blown up inside a cylinder, and sticks to the cylinder when it comes into contact with it.
unroll the cylinder to get the map
Two properties:
circles of latitude become horizontal lines on the map
meridians become vertical lines
Result: latitudes and longitude lines form a grid pattern
Contest Algorithms:16. Lat & Long

35. Also called a
loxodrome
A constant course always
make the same angle
with the meridians (longitude line) that
it passes through.
constant course == a rhumb line on a Mercator map

36. Calculating Rhumb Line Info
Derive the equations to calculate the Rhumb line course C and its distance d between two points A and B.
d can be calculated in terms of a sum of small distances dx between A and B:
dLon * cos Lat
lattitude
lines
divide d into
lots of small triangles d dx
dLat C C A
meridians
Contest Algorithms

37. dx is composed of a north-south component, dLat, and a west-east component, dLon ∙ cos Lat.
The factor (cos Lat) is the relative circumference of the respective parallel of latitude.
Rearrange:
Contest Algorithms:16. Lat & Long

38. Sum all the dx triangles by integrating the equations between (Lat A, Lat B) and (Lon A, Lon B):
1/cos θ = sec θ
sec θ = ln( | sec θ + tan θ |) + K
= tan(θ/2 + π/4) + K
Contest Algorithms:16. Lat & Long

39. Integrate both sides:
Rearrange:
Contest Algorithms:16. Lat & Long

40. Solve for C, and use atan2(y,x) rather than arctan()
C = Math.atan2( , )
Contest Algorithms:16. Lat & Long

41. To find d, sum the dx lengths, by integrating the dx equation:
Result:
𝐷= 𝐿𝑎𝑡 𝐵 − 𝐿𝑎𝑡 𝐴 cos 𝐶
Contest Algorithms:16. Lat & Long

42. Rhumb Line from NY to Moscow
Coordinates:
New York: Latitude = 40o47'N, Longitude = 73o58'W
Moscow: Latitude = 55o45'N, Longitude = 37o42'E
Calculation:
diffLat = = o
diffLon = = o = radians
tan(LatMO/2 + π/4) = ; tan(LatNY/2 + π/4) =
ln( / ) =
C = atan2(1.949, ) = radians = 78.5o
Contest Algorithms:16. Lat & Long

43. D = / cos(1.37) = / =
NM distance == * 60 = NM
km distance = * = km
Contest Algorithms:16. Lat & Long

44. Online Rhumb Line Calculator
Contest Algorithms:16. Lat & Long

45. Code
public static double rhumbDistance(double lat1d, double lon1d, double lat2d, double lon2d) // return distance in km { double lat1 = Math.toRadians(lat1d); double lon1 = Math.toRadians(lon1d); double lat2 = Math.toRadians(lat2d); double lon2 = Math.toRadians(lon2d); if ((Math.abs(lat1 - lat2) < EPS) && (Math.abs(lon1 - lon2) < EPS)){ System.out.println("Too points are the same"); return 0; } double dPhi = Math.log(Math.tan(lat2/2.0 + Math.PI/4.0) / Math.tan(lat1/2.0 + Math.PI/4.0)); double course = Math.atan2((lon2 - lon1), dPhi); System.out.printf("Rhumb line course (in degrees): %.4f\n", Math.toDegrees(course)); double distNM = ((lat2d - lat1d) / Math.cos(course) * 60); // in Nautical miles return (distNM * NAUTICAL_MILES_TO_KM); } // end of rhumbDistance()
Contest Algorithms:16. Lat & Long

46. Usage
// in main() System.out.printf("Rhumb line distance (in km): %.1f\n", rhumbDistance(nyLat, nyLong, mLat, mLong) );
Contest Algorithms:16. Lat & Long