Exo Cruiser: Extreme Points of a Great Circle

The Northernmost and Southernmost Points of a Great Circle

If you travel from Amsterdam (P1 in Figure 1) to San Francisco (P2) or the other way around, then you first go towards the north for a while, and then towards the south for a while. All great circles except for the equator have a northernmost point (PN) and a southernmost point (PS). You can calculate them as follows.

Fig. 1: Northernmost and Southernmost Point on a Great Circle

Repeating here what we already had in part 1 calculate the Cartesian coordinates of both points P1 and P2.

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 (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

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

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

3. 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):

Fig. 3. P3 is 90 degrees from P1 towards P2.

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.

4. Calculate the angular distance of the first city P1 from the first special (northernmost or southernmost) point:

phi1 = atan(z3/z1) (11)

The angular distance of the second special point is 180° greater (or less, that is the same thing on a circle):

phi2 = phi1 + 180° (12)

For the great circle that passes through Amsterdam and San Francisco, we find

phi1 = 30.62449°;
phi2 = 210.62449°.

5. You can then use formula 6 to calculate the corresponding cartesian coordinates, and then formula 7ff to calculate the polar coordinates as in part 1.

The corresponding cartesian coordinates are as follows (if you get the sothernmost instead of the northernmost point then swap the angles phi1 and phi2)

xn = x1*cos(phi1) + x3*sin(phi1) (6)
yn = y1*cos(phi1) + y3*sin(phi1)
zn = z1*cos(phi1) + z3*sin(phi1)

xn = 0.2427036;
yn = -0.3067243;
zn = 0.9203343

The corresponding polar coordinates are

bn = asin(zn) (7)
ln = atan2(yn,xn) (8) (see note *1)

bn = 66.975°;
ln = -51.64627°

for phi1 (northernmost) and

xs = x1*cos(phi2) + x3*sin(phi2) (6)
ys = y1*cos(phi2) + y3*sin(phi2)
zs = z1*cos(phi2) + z3*sin(phi2)

xs = -0.2427036;
ys = 0.3067243;
zs = -0.9203343

bs = asin(zs) (7)
ls = atan2(ys,xs) (8)

bs = -66.975°;
ls = 128.3537°

for (southernmost) phi2.

It is not necessary to calculate the coordinates of the second special point, because it is at the exact opposite side of the planet from the first one, so its Cartesian coordinates and its latitude are equal to those of the first point, times -1, and its longitude is 180° around the planet from the first special point.

Fig. 8. Gerardus Mercator the creator of Mercator projection.

["Gerardus Mercator (5 March 1512 – 2 December 1594) was a 16th-century German-Flemish cartographer, geographer and cosmographer. He was renowned for creating the 1569 world map based on a new projection which represented sailing courses of constant bearing (rhumb lines) as straight lines—an innovation that is still employed in nautical charts."]

The Equator Points of a Great Circle

Every great circle except for the equator intersects the equator in two points, called E1 and E2 in Figure 1. The longitudes of those points are 90° to the east and west of the northernmost and southernmost points of the great circle, of which the calculation is explained above.

Calculate E1 as follows:

phi3 = phi1 + 90 = 120.624489505494680

xe1 = x1*cos(phi3) + x3*sin(phi3)=-0.784194876815035
ye1 = y1*cos(phi3) + y3*sin(phi3)=-0.620514621243571
ze1 = z1*cos(phi3) + z3*sin(phi3) = 0

be1 = asin(ze1) = 0
le1 = atan2(ye1,xe1) = -141.646275410932280

Calculate E2 as follows:

phi4 = phi1 + 270 = 300.624489505494690

xe2 = x1*cos(phi4) + x3*sin(phi4) =0.784194876815035
ye2 = y1*cos(phi4) + y3*sin(phi4) =0.620514621243571
ze2 = z1*cos(phi4) + z3*sin(phi4) =0

be2 = asin(ze2) = 0
le2 = atan2(ye2,xe2) = 38.353724589067724

So, the great circle through Amsterdam and San Francisco crosses the equator at longitudes 38.3537° and -141.64627°, in Kenya and the Pacific Ocean, respectively.

Fig. 4. The extreme points (northernmost, southernmost, easternmost and westernmost) shown on Robinson projection. Notice how the great circle acts strange when behind the sphere.

[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).]

Conclusions

It is interesting how some large maps give us wrong ideas. For example looking the following Mercator projection map would give you the idea that the Rhumb line from Amsterdam to San Francisco is the best when it actually is 1300 km longer than the Great circle.