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## 1. A Great Circle

Fig. 1: Drawing of a Great Circle on a Sphere
A great circle is a special circle at the surface of a sphere (for example of a planet or of the sky). 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 divides the surface of a sphere in two equal 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.
• A great circle has the same center C as the sphere that it lies on.
• 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.

## 2. 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:

1. Call the polar coordinates (longitude and latitude) of the first city $$l_1$$ and $$b_1$$, and those of the second city $$l_2$$ and $$b_2$$.
2. Fig. 2: Illustration of Transformation from Polar to Cartesian Coordinates
Translate the polar coordinates of the first city P₁ to the corresponding cartesian coordinates $$x_1$$, $$y_1$$, $$z_1$$ (see figure 2):

\begin{align} x_1 \| = \cos l_1 \cos b_1 \label{eq:naarcartesisch} \\ y_1 \| = \sin l_1 \cos b_1 \\ z_1 \| = \sin b_1 \end{align}

and similarly for the second city P₂.

3. Fig. 3: Illustration of the Angular Distance Between P₁ and P₂
Calculate the angular distance $$ψ$$ (psi) between the two cities, as seen from the center of the Earth (see Figure 3):

$$ψ = \arccos(x_1 x_2 + y_1 y_2 + z_1 z_2)$$

4. 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):

$$x_3 = \frac{x_2 - x_1 \cos ψ}{\sin ψ} \label{eq:punt3}$$

and similarly with $$y$$ or $$z$$ in stead of $$x$$.

5. The cartesian coordinates of the points of the great circle are then, as a function of the angular distance $$φ$$ (phi) from the first city:

$$x = x_1 \cos φ + x_3 \sin φ \label{eq:positie}$$

and similarly with $$y$$ or $$z$$ in stead of $$x$$. If $$φ = 0$$, then you are in the first city. If $$φ = ψ$$, then you are in the second city.

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

\begin{align} b \| = \arcsin(z) \label{eq:naarpolair} \\ l \| = \arctan(y,x) \end{align}

The $$\arctan(y,x)$$ with two arguments means that you must make sure that the answer is in the right quadrant. The correct answer is either $$\arctan\left( \frac{y}{x} \right)$$, or $$\arctan\left( \frac{y}{x} \right) + 180°$$, 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.

Fig. 4: Great Circle Through Amsterdam and San Francisco
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? The distance per degree across the sphere is equal to the radius times $$π/180 = 0.017453292$$, so on Earth this is 111.317 km per degree. We find:

1. $$l_1 = 4.9°$$; $$b_1 = 52.37°$$; $$l_2 = −122.42°$$; $$b_2 = 37.77°$$
2. $$x_1 = 0.6083285$$; $$y_1 = 0.05215215$$; $$z_1 = 0.7919701$$; $$x_2 = −0.423791$$; $$y_2 = −0.6672729$$; $$z_2 = 0.6124933$$
3. $$ψ = 78.90289°$$, so San Francisco is $$78.90289 × 111.317 = 8783$$ km from Amsterdam.
4. $$x_3 = −0.5511833$$; $$y_3 = −0.6902162$$; $$z_3 = 0.4688268$$. This corresponds to $$b_3 = 27.95817°$$; $$l_3 = −128.6097°$$, which is a location in the Eastern Pacific Ocean, to the West of Mexico.
5. 1000 km corresponds to $$1000/111.317 = 8.98335$$ degrees, so $$φ = 8.98335°$$. Then $$x = 0.5148008$$; $$y = −0.05626304$$; $$z = 0.8554617$$.
6. $$b = 58.81077°$$; $$l = −6.237153°$$. This is a location just to the North of Scotland.

## 3. Alternative

There is an alternative for formula \ref{eq:positie}, which does not require the calculation of the position of point 3:

$$x = \frac{x_1 \sin(ψ - φ) + x_2 \sin φ}{\sin ψ}$$

and similarly with $$y$$ or $$z$$ instead of $$x$$. However, point 3 is necessary if you want to know other things about the great circle, as you'll see below.

## 4. A Great Circle in a Certain Direction Through a Known Point

Suppose you want to know where you go if you start from a particular town in a particular direction and keep going straight. We assume that you travel at a speed that is much smaller than the rotation speed of the Earth along the equator. You'll then travel along a great circle. You can calculate the coordinates of points along the route as follows:

1. Call the polar coordinates (longitude and latitude) of the city $$l_1$$ and $$b_1$$, and the direction in which you start $$γ$$, measured from south to west (so south = 0°, west = 90°, north = 180°, east = 270°).
2. Translate the polar coordinates of the city to the corresponding cartesian coordinates $$x_1$$, $$y_1$$, $$z_1$$ according to formulas \ref{eq:naarcartesisch}ff.
3. Calculate the cartesian coordinates $$x_\text{zuid}$$, $$y_\text{zuid}$$, $$z_\text{zuid}$$ of the corresponding south point with $$l_\text{south} = l_1$$; $$b_\text{south} = b_1 - 90°$$ if $$b_1$$ is positive (i.e., in the Northern hemisphere), and $$l_\text{south} = l_1 + 180°$$; $$b_\text{south} = −90° - b_1$$ if $$b_1$$ is negative (i.e., in the southern hemisphere).
4. Calculate the cartesian coordinates $$x_\text{west}$$, $$y_\text{west}$$, $$z_\text{west}$$ of the corresponding west point with $$l_\text{west} = l_1 - 90°$$; $$b_\text{west} = 0$$.
5. Calculate the cartesian coordinates $$x_3$$, $$y_3$$, $$z_3$$ of the great circle point at 90° from the city:

$$x_3 = x_\text{south} \cos γ + x_\text{west} \sin γ$$

and similarly with $$y$$ or $$z$$ instead of $$x$$.

6. Now use formulas \ref{eq:positie} and \ref{eq:naarpolair}ff to calculate the desired cartesian and polar coordinates.

Suppose you start from Amsterdam (52°22' north, 4°54' east) by going straight to the east, and you keep going straight. If you keep this up for 1000 km, then where are you?

1. $$l_1 = 4.9°$$; $$b_1 = 52.37°$$; $$γ = 270°$$
2. $$x_1 = 0.6083285$$; $$y_1 = 0.05215215$$; $$z_1 = 0.7919701$$
3. $$x_\text{south} = 0.7890756$$; $$y_\text{south} = 0.06764765$$; $$z_\text{south} = −0.6105599$$
4. $$x_\text{west} = 0.08541692$$; $$y_\text{west} = −0.9963453$$; $$z_\text{west} = 0$$
5. $$x_3 = −0.08541692$$; $$y_3 = 0.9963453$$; $$z_3 = 0.7919701$$, so in this case each cartesian coordinate of point 3 is the opposite of the corresponding coordinate of the west point, which was to be expected because we start out going straight to the east.
6. 1000 km corresponds to 8.98335°, so $$φ = 8.98335°$$. With that, we find $$x = 0.587529$$; $$y = 0.2070892$$; $$z = 0.7822556$$, en dan $$b = 51.46756°$$; $$l = 19.41627°$$. This is a location in the middle of Poland.

## 5. The Northernmost and Southernmost Point of a Great Circle

Fig. 5: Northernmost and Southernmost Point on a Great Circle
If you travel from Amsterdam (P₁ in Figure 5) 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, if you have points P₁ and P₃ separated by 90° along the great circle (for example, by using formula \ref{eq:punt3}):

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

$$φ_1 = \arctan\left( \frac{z_3}{z_1} \right)$$

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

$$φ_2 = φ_1 + 180°$$

You can then use formula \ref{eq:positie} to calculate the corresponding cartesian coordinates, and then formula \ref{eq:naarpolair}ff to calculate the polar coordinates. 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.

For the great circle that passes through Amsterdam and San Francisco, we find $$φ_1 = 30.62449°$$; $$φ_2 = 210.62449°$$. The corresponding cartesian coordinates are $$x = −0.2427036$$; $$y = 0.3067243$$; $$z = −0.9203343$$ for $$φ_1$$, and $$x = 0.2427036$$; $$y = −0.3067243$$; $$z = 0.9203343$$ for $$φ_2$$. The corresponding polar coordinates are $$b = 66.975°$$; $$l = −51.64627°$$ for $$φ_1$$ and $$b = −66.975°$$; $$l = 128.3537°$$ for $$φ_2$$.

## 6. Distance from a Great Circle

What is the distance from an arbitrary point Q to the great circle through P₁ and P₂, measured along the surface perpendicular to the great circle?

The short answer is that that distance $$α$$ is determined by

$$\sin(α) = \frac{\hat{Q} ⋅ (\hat{P}_1 × \hat{P}_2)}{|\hat{P}_1 × \hat{P}_2|} \label{eq:dist}$$

where $$⋅$$ indicates the inner product of two vectors, $$×$$ the vector cross product of two vectors, and $$|\vec{v}|$$ indicates the length of vector $$\vec{v}$$. The notation $$\hat{v}$$ says that vector $$\vec{v}$$ has a length equal to 1.

Here is the same answer, but with more explanation. The vectors $$\hat{P}_1$$, $$\hat{P}_2$$, and $$\hat{Q}$$ point from the center of the sphere to the points P₁, P₂, and Q. Their Cartesian coordinates can be found from their polar coordinates using Eq. \eqref{eq:naarcartesisch}.

Vector $$\vec{S} = \hat{P}_1 × \hat{P}_2$$ by definition makes a right angle with both $$\hat{P}_1$$ and $$\hat{P}_2$$, but its length is only equal to 1 if $$\hat{P}_1$$ and $$\hat{P}_2$$ are at a right angle themselves. Because its length might not be equal to 1, we write that vector as $$\vec{S}$$ and not as $$\hat{S}$$.

The inner product of $$\hat{Q}$$ and $$\vec{S}$$ is by definition related to the angle $$β$$ between them, through

$$|\hat{Q}| |\vec{S}| \cos(β) = \hat{Q} ⋅ \vec{S}$$

but because the length of $$\hat{Q}$$ is $$|\hat{Q}| = 1$$, we don't need to write that factor explicitly, so

$$|\vec{S}| \cos(β) = \hat{Q} ⋅ \vec{S}$$

from which follows

$$\cos(β) = \frac{\hat{Q} ⋅ \vec{S}}{|\vec{S}|}$$

Because $$\vec{S}$$ is perpendicular to (the plane of) the great circle, the sum of the angle between $$\hat{Q}$$ and $$\vec{S}$$ and the angle between $$\hat{Q}$$ and (the plane of) the great circle is 90°, so

\begin{eqnarray} β \| = \| 90° − α \\ \cos(β) \| = \| \sin(α) \end{eqnarray}

That leads us to Eq. \eqref{eq:dist}. The sign of $$\sin(α)$$ depends on the order of P₁ and P₂. If we swap P₁ and P₂, then $$\sin(α)$$ and $$\vec{S}$$ get multiplied by −1. The $$z$$ coordinate of $$\vec{S}$$ says whether $$\vec{S}$$ points to a location in the northern ($$z_S \gt 0$$) or southern ($$z_S \lt 0$$) hemisphere. If $$z_S\sin(α) \gt 0$$, then Q lies on the north side of the great circle, and if $$z_S\sin(α) \lt 0$$, then Q lies on the south side of the great circle.

It remains to explain that the inner product of two vectors is equal to

$$\Matrix{x_1 \\ y_1 \\ z_1} ⋅ \Matrix{x_2 \\ y_2 \\ z_2} = x_1x_2 + y_1y_2 + z_1z_2$$

and the vector cross product is equal to

$$\Matrix{x_1 \\ y_1 \\ z_1} × \Matrix{x_2 \\ y_2 \\ z_2} = \Matrix{y_1z_2 − y_2z_1 \\ z_1x_2 − z_2x_1 \\ x_1y_2 − x_2y_1}$$

For example, how far from the great circle through Amsterdam (P₁) and San Francisco (P₂) is the airport of Reykjavik in Iceland (Q, at 64.13° north lattiude and 21.94° west longitude)?

We found before that for Amsterdam $$x_1 = 0.6083285$$; $$y_1 = 0.05215215$$; $$z_1 = 0.7919701$$ and for San Francisco $$x_2 = −0.423791$$; $$y_2 = −0.6672729$$; $$z_2 = 0.6124933$$.

For Reykjavik we find, using Eq. \eqref{eq:naarcartesisch}, the Cartesian coordinates $$x_Q = 0.4047297$$; $$y_Q = −0.1630286$$; $$z_Q = 0.8997864$$.

Then

\begin{eqnarray*} \vec{S} \| = \| \Matrix{0.6083285 \\ 0.05215215 \\ 0.7919701} × \Matrix{−0.423791 \\ −0.6672729 \\ 0.6124933} = \Matrix{0.560403 \\ −0.7082269 \\ −0.3838195} \\ |\vec{S}| \| = \| \sqrt{0.560403^2 + (−0.7082269)^2 + (−0.3838195)^2} = \sqrt{0.9629544} = 0.9813024 \\ \hat{Q} ⋅ \vec{S} \| = \| 0.4047297 × 0.560403 + (−0.1630286) × (−0.7082269) + 0.8997864 × (−0.3838195) = −0.00308254 \\ \sin(α) \| = \| \frac{−0.00308254}{0.9813024} = −0.003141275 \\ α \| = \| \arcsin(−0.003141275) = −0.1799821° \end{eqnarray*}

We find that $$z_S\sin(α) \gt 0$$, so Reykjavik is to the north of the great circle through Amsterdam and San Francisco, at a distance of 0.18°.

## 7. The Equator Points of a Great Circle

Every great circle except for the equator intersects the equator in two points, called E_1 and E_2 in Figure 5. 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.

For example, 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.

## 8. Great Circle on a Topographic Map

Fig. 6: Great Circle on a Map
Figure 6 shows all relevant points and the relevant great circle from the above examples on a world map. This type of world map is often used but is not very suitable for finding the shortest route (part of a great circle) between two points on Earth, because most great circles appear crooked on such maps, just like the great circle that is shown.

It is impossible to make a map of the world on which all great circles run straight, but it is possible to make a map on which some great circles run straight, for example all great circles through a single point. In Figure 6, all great circles through the North Pole and South Pole run straight: Those are the meridians. The equator is a great circle and it, too, runs straight through that map.

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Last updated: 2021-07-19