Exo Cruiser: Definition of a Great Circle

[Since great circle calculations are so important on all spheres (like Earth, planets, polar coordinates, etc.) I have to add here such a text. This follows mostly /1/]

Fig. 7. Arthur H. Robinson 1979

["Arthur H. Robinson (January 5, 1915 – October 10, 2004) was an American geographer and cartographer, who was professor in the Geography Department at the University of Wisconsin–Madison from 1947 until he retired in 1980. He was a prolific writer and influential philosopher on cartography.

One of Robinson's most notable accomplishments is the Robinson projection. In 1961, Rand McNally asked Robinson to choose a projection for use as a world map that, among other criteria, was uninterrupted,[9] had limited distortion, and was pleasing to the eye of general viewers.[10] Robinson could not find a projection that satisfied the criteria, so Rand McNally commissioned him to design one.

Robinson proceeded through an iterative process to create a pseudo-cylindrical projection that intends to strike a compromise between distortions in areas and in distances, in order to attain a more natural visualization. The projection has been widely used since its introduction. In 1988, National Geographic adopted it for their world maps but replaced it in 1998 with the Winkel tripel projection."]

A Great Circle

A great circle is the largest circle you can draw on a sphere, the equator for example. It is a special circle at the surface of a sphere (for example of a planet or of the sky).

Fig. 1: Drawing of a Great Circle on a Sphere

Figure 1 shows great circle G and latitude circle B on a sphere. You can recognize a great circle by any of the following attributes:

A great circle's center is always also the center of the sphere. All great circles meet at the center of the sphere.
A great circle divides the surface of a sphere in two exactly equal size parts.
A great circle is the largest circle that fits on the sphere.
If you keep going straight across a sphere then you go along a great circle.
The shortest route between two points, measured across the sphere, is part of a great circle.

All meridians are great circles. Latitude circles other than the equator (for example circle B in the picture) are not great circles, for example because they are smaller than the equator, which is a great circle.

A great circle provides the shortest route if you travel at a fixed speed compared to the ground, and also (approximately) if your speed, though not fixed, is always much smaller than the rotation speed of the sphere at its equator. This does not hold, for example, for things that orbit around the Earth outside the atmosphere.

For example an airplane will always use least energy between two points if it follows the great circle between these two points what ever the winds might be. A sail ship might sometimes be a different question since it uses its keel but usually not.

[If you wish to verify the following calculations use xCalc for example to do it (it is a free application which can be used to calculate formulas in Windows)]

Find a Great Circle Through Two Known Points

Suppose that you want to draw the shortest route on a map between a city P1 and a far-away other city P2, and you know the geographical longitude and latitude of both cities. Then, you can calculate the coordinates of points on that route as follows.

Fig. 4: Great Circle Through Amsterdam P1 and San Francisco P2

For example (Figure 4): Which point lies 1000 km from Amsterdam (P1, 52°22' North, 4°54' East) on the shortest route to San Francisco (P2, 37°46' North, 122°25' West), assuming that the Earth is a sphere with a radius of 6378 km?

Figure 6. Great Circle arc from Amsterdam to San Francisco

1. Call the polar coordinates (longitude and latitude) of the first city l1 and b1, and those of the second city l2 and b2.

l1 = 4.9°; b1 = 52.37°;
l2 = -122.42°; b2 = 37.77°

2. Translate the polar coordinates of the first city P1 to the corresponding Cartesian coordinates x1, y1, z1 on a unit sphere (see Figure 2):

Fig. 2: Illustration of Transformation from Polar to Cartesian Coordinates

  x1 = cos(l1)*cos(b1)    (1)
  y1 = sin(l1)*cos(b1) (2)
  z1 = sin(b1) (3)

  x1=0.6083285;
  y1=0.05215215;
  z1=0.7919701;

and similarly for the second city P2.

x2 = cos(l2)*cos(b2)
  y2 = sin(l2)*cos(b2)
  z2 = sin(b2)

  x2 = -0.423791;
  y2 = -0.6672729;
  z2 = 0.6124933

3. Calculate the angular distance psi between the two cities, as seen from the center of the Earth (see Figure 3):

Fig. 3: Illustration of the Angular Distance Between P1 and P2

  psi = acos(x1*x2 + y1*y2 + z1*z2)    (4)
  psi = 78.90289°

The distance per degree u across the sphere is equal to

  u = r*pi/180, where r = 6378 km
  u = 111.317 km/°

so on Earth this is 111.317 km per degree. So San Francisco is

  d = psi*u
  d = 78.90289*111.317 = 8783 km

from Amsterdam.

4. To define the great circle for the purpose of other calculations calculate the coordinates of the point P3 on the great circle that is 90° from the first city P1 in the direction of the second city P2 (see Figure 3):

  x3 = (x2 - x1*cos(psi))/sin(psi)    (5)
  y3 = (y2 - y1*cos(psi))/sin(psi)
  z3 = (z2 - z1*cos(psi))/sin(psi)

  x3 = -0.5511833;
  y3 = -0.6902162;
  z3 = 0.4688268.

This corresponds to

  b3 = asin(z3) = 27.95817°;
  l3 = atan2(y3,x3) = -128.6097°, (see note *1)

which is a location in the Eastern Pacific Ocean, to the West of Mexico.

Fig. 6. P3 is located west of Mexico (Robinson projection).

5. Since dist=1000 km corresponds to an angle phi of

  phi = dist/u
  phi = 8.98335°

we can now calculate the point in question.

The cartesian coordinates of the points of the great circle are then, as a function of the angular distance phi from the first city:

  x = x1*cos(phi) + x3*sin(phi)    (6)
  y = y1*cos(phi) + y3*sin(phi)
  z = z1*cos(phi) + z3*sin(phi)

  x = 0.5148008;
  y = -0.05626304;
  z = 0.8554617.

If phi = 0, then you are in the first city (Amsterdam). If phi = psi, then you are in the second city (San Francisco). If phi = 8.98335° we are 1000 km from the first point towards the second point. If phi = 90° then we are at the P3 which defines our great circle in this question.

6. You can now translate the Cartesian coordinates x, y, z to polar coordinates l, b:

  b = asin(z) (7)
  l = atan2(y,x)    (8)

  b = 58.81077°;
  l = -6.237153°.

This is a location just to the North of Scotland.

Fig 5. Result P is located north of Scotland on a Robinson projection. Notice how all map projections distort long great circle arcs. If your distance is longer than 500 km it is a very good idea to check if the shortest distance line is any close to a line at all. The above curve is the shortest way from Amsterdam to San Francisco and to do it you have to fly over Greenland.

[Note *1:

The atan2(y,x) with two arguments means that you must make sure that the answer is in the right quadrant. The correct answer is either atan(y/x), or atan(y/x)+180°, and (in this case) you must select the solution that has x for its cosine and y for its sine (with the correct signs).

Many computer languages and computer calculation programs have a two-argument version of the arc tangent function, and many calculators have a translation function from cartesian to polar coordinates that you can use for this. For example xCalc for Windows can do all these calculations (download here).]

Alternative

There is an alternative for formula 6, which does not require the calculation of the position of point 3:

  x = (x1*sin(psi-phi) + x2*sin(phi)) / sin(psi)    (9)
      =0.514800782206212
  y = (y1*sin(psi-phi) + y2*sin(phi)) / sin(psi)
      =-0.056262996832530
  z = (z1*sin(psi-phi) + z2*sin(phi)) / sin(psi)
      =0.855461647198339

However, point P3 is necessary if you want to know other things about the great circle, as you'll see in the following parts of this text.