## The great circle distance

### How can we find the formula for the great circle distance?

In Lost but lovely: the haversine we explored great circle distances on the Earth and looked at the haversine formula for calculating them:

In this piece, we explore where this formula comes from, and why navigators might have used the unfamiliar haversine rather than ordinary sines and cosines in their calculations.

Given a point on the unit sphere with latitude *ϕ* and longitude *λ*, we can calculate the corresponding Cartesian coordinates, as shown in the following diagram.

What are the Cartesian coordinates of *P* (that is, its (*x*,*y*,*z*) coordinates)?

Once we know the Cartesian coordinates of the points *P* and *Q* with latitudes *ϕ*1 and *ϕ*2 and longitudes *λ*1 and *λ*2 respectively, we can calculate the differences in the *x*, *y* and *z* coordinates of the two points and use those to calculate appropriate distances or angles in order to find *d*.

However, the expressions are going to become quite cumbersome with so many variables, so we look for a way to simplify things before going further.

One trick is to notice that the distances and angle *σ* all remain unchanged if we rotate the sphere about the north-south axis. This only changes the longitudes of the points, so we could rotate it until *P* has longitude 0 and *Q* has longitude *λ*2−*λ*1, which we will call Δ*λ* (the difference between the *λ*s).

Then the coordinates of *P* and *Q* become:

We can then continue exactly as in Approach 1.

#### Approach 3: Spherical trigonometry

Just as there are sine and cosine rules for triangles in the plane, there are similar rules for triangles on a sphere.

Here is a drawing of a spherical triangle:

The angles *A*, *B* and *C* are the angles at the vertices *A*, *B* and *C* of the triangle, while *a*, *b* and *c* are the side lengths as usual. But if we take the sphere to have radius 1, then they can also be thought of as angles: *a* is the angle ∠*B**O**C*, for example. We therefore assume the sphere to have radius 1

from now on.

There are two spherical cosine rules and one spherical sine rule, which are:

- spherical cosine rule (1):

In this diagram, the blue lines are all straight chords (though an optical illusion makes it look otherwise), and the red line is the great circle between *P* and *Q*

.Can you use this diagram to deduce the haversine formula?

The first thing to observe is that the lines *PR* and *SQ* are parallel. This is because the circles at latitudes *ϕ*1 and *ϕ*2 are just scaled versions of each other. This means that the quadrilateral *PRQS* is actually planar and a trapezium. Furthermore, the lengths *PS* and *RQ* are the same by construction. The trapezium is therefore an isosceles trapezium. We can redraw it like this:

How do these four approaches compare?

#### Why use haversines?

Surely we could just use cos(* ^{d}/_{R}*)=cos

*ϕ*1cos

*ϕ*2cos(Δ

*λ*)+sin

*ϕ*1sin

*ϕ*2 to calculate great circle distances, as we found in Approaches 2 and 3? Why would anyone need the haversine? (We have reinserted the radius

*R*here, as practical calculations will use standard units of measurement. Also, a reason is given in Lost but lovely: the haversine for introducing the haversine, but that was just to simplify the formula involving three occurrences of sin

^{2}

*x*/

_{2}.)

Navigators in the past would have had to make do with tables and slide rules: there were no calculators. So what would happen if we try to work out the distance between Dover and Calais, say, using the two different formulae, but restricting ourselves to the limited calculating tools of the past?

Here is some relevant information:

- Dover: latitude 51.15∘N

, longitude 1.33∘E Calais: latitude 50.97∘N, longitude 1.85∘E Radius of the Earth: 6378km

And trigonometric values, accurate to 4 significant figures or 4 decimal places:

Using these values, calculate the great circle distance from Calais to Dover using the two different formulae (and you can use a calculator for the multiplications, and for working out the final inverse cosine and haversine). To truly emulate the past, round all of your intermediate calculations to 4 significant figures as well.

small, then sin(Δ*λ*)≈Δ*λ* (as long as we are working in radians), so working to 4 significant figures will still be very accurate.

## Great Circle Formula

The great circle is the largest circle that can be drawn on the surface of a sphere. A great circle is a section of a sphere that contains the diameter of the sphere and the diameter is the shortest distance between any two points on the sphere surface is the great circle distance. The great circle is also known as Romanian Circle. One of the few examples where the application of the great circle formula is used is in the navigation of aircraft or ships. As we know the Earth is somewhat spherical, the great circle formula helps in navigating as we come to know for the shortest distance in the sphere.

## What is the Great Circle Formula?

Great circle formula can be expressed as,

d=rcos−1[cosacosbcos(x−y)+sinasinb]

where,

- r = Radius of the earth
- a, b= Latitude
- x, y = Longitude

Note: Sphere’s diameter coincides with the diameter of the great circle.

Let’s take a quick look at a couple of examples to understand the great circle formula, better.

## Examples Using Great Circle Formula

**Example 1: **What will be the length of the great circle if the radius of the sphere is 5 km, the latitude is (25^{o}, 34^{o}) and the longitude is (48^{o}, 67^{o}).

**Solution:**

To find: Great circle distance.

Given:

r = 5 km

a = 25^{o} = 0.418 rad, and b = 34^{o}= 0.59 rad

x = 48^{o }= 0.83 rad, y = 67^{o} = 1.16 rad

Using great circle formula,

d=rcos−1[cosacosbcos(x−y)+sinasinb]d=5×cos−1[0.913×0.83×0.945+0.406×0.556]=5×2.793

**Answer: **Hence, the great circle length is 94.928 km.

**Example 2: **What will be the length of the great circle if the radius of the sphere is 10 km, the latitude is (55^{o}, 86^{o}) and the longitude is (28^{o}, 70^{o}).

**Solution:**

To find: Great circle distance.

Given:

r = 10 km

a = 55^{o},= 0.959 rad, b = 86^{o}= 1.5

x = 28^{o}= 0.48 y = 70^{o} = 1.22

Using Great Circle Formula,

d=rcos−1[cosacosbcos(x−y)+sinasinb]d=10×cos−1[0.573×0.0707×0.743+0.819×0.997]=10×0.562

**Answer: **Hence, the great circle distance is 5.62 km.

## Great Circle

A great circle is a section of a sphere that contains a diameter of the sphere (Kern and Bland 1948, p. 87). Sections of the sphere that do not contain a diameter are called small circles. A great circle becomes a straight line in a gnomonic projection (Steinhaus 1999, pp. 220-221).

## How to Calculate and Solve Distance Along Great Circles | Latitude and Longitude

The image above represents distance along great circles.

To compute for distance along great circles, two essential parameters are needed and these parameters are **value of R **and **angular difference (θ).**

The formula for calculating the distance along great circles:

y = (^{θ} / _{360}) x 2πR

Where;

y = Distance Along Great Circles

θ = Angular Difference

R = Radius of the Earth

Let’s solve an example;

Find the distance along great circles when the value of R is 6400 and the angular difference is 18.

This implies that;

θ = Angular Difference = 18

R = Radius of the Earth = 6400

y = (^{θ} / _{360}) x 2πR

y = (^{18} / _{360}) x (2 x π x 6400)

y = 0.05 x 40212.38

y = 2010.6

Therefore, the **distance along great circles **is **2010.6 km.**

**Calculating the Angular Difference when the Distance Along Great Circles and the Value of R is Given.**

θ = ^{y360} / _{2πR}

Where;

θ = Angular Difference

y = Distance Along Great Circles

R = Radius of the Earth

Let’s solve an example;

Given that the distance along great circles is 32 and the value of R is constant (6400). Find the angular difference?

This implies that;

y = Distance Along Great Circles = 32

R = Radius of the Earth = 6400

θ = ^{y360} / _{2πR}

θ = ^{(32)360} / _{2π(6400)}

θ = ^{11520} / _{6.28(6400)}

θ = ^{11520} / _{40192}

θ = 0.286

Therefore, the **angular difference **is **0.286°.**

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