Great circle distance

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Great circle distance is the shortest distance between any two points on a sphere.

Because the Earth is approximately spherical, the great circle distance is commonly used to find the distance (e.g. in miles) between two coordinates (latitude/longitude pairs) on a map.

Table of contents

1 Definitions

2 The formula

3 Practical application

3.1 Choose two points
3.2 Convert coordinates
3.3 Solve

4 Quality of the approximation

5 See also

Definitions

<math>R<math> is the radius of the Earth (<math>\approx<math> 6367000 meters)

<math>\delta<math> is latitude

<math>\lambda<math> is longitude

The formula

<math>\operatorname{gc}(\delta, \lambda, \delta', \lambda') <math>

<math>=2R \arcsin\sqrt{\sin^2{\left(\frac{\delta' - \delta}{2}\right)} + \cos{\delta} \cdot \cos{\delta'} \cdot \sin^2{\left(\frac{\lambda' - \lambda}{2}\right)}\ }<math>

Practical application

Choose two points

In order to use this formula for anything practical you will need two sets of coordinates. For example, the latitude and longitude of two airports:

Nashville International Airport (BNA) in Nashville, TN, USA: N 36°7.2', W 86°40.2
Los Angeles International Airport (LAX) in Los Angeles, CA, USA: N 33°56.4', W 118°24.0'

Convert coordinates

You will have to convert these coordinates to a more mathematically friendly form using a simple method before you can use them effectively in a formula. After conversion, the coordinates become:

BNA: <math>\delta<math> = 36.12�, <math>\lambda<math> = −86.67�
LAX: <math>\delta'<math> = 33.94�, <math>\lambda'<math> = −118.40�

You'll need to convert these coordinates to radians instead of degrees for them to be useful in the formula:

BNA: <math>\delta<math> = 0.630412925820352, <math>\lambda<math> = −1.5126768627035
LAX: <math>\delta'<math> = 0.59236474812688, <math>\lambda'<math>= −2.0664698343613

Solve

Now you can simply substitute numbers in the formula above. This is best done in small steps to keep each step manageable:

<math>\operatorname{gc}(\delta, \lambda, \delta', \lambda') <math>

<math>=2R \arcsin\sqrt{\sin^2{\left(\frac{\delta' - \delta}{2}\right)} + \cos{\delta} \cdot \cos{\delta'} \cdot \sin^2{\left(\frac{\lambda' - \lambda}{2}\right)}\ }<math>

<math>=2R \arcsin\sqrt{\sin^2{\left(\frac{0.59236474812688 - 0.630412925820352}{2}\right)} + \cos{0.630412925820352} \cdot \cos{0.59236474812688} \cdot \sin^2{\left(\frac{-2.0664698343613 - -1.5126768627035}{2}\right)}\ }<math>

<math>=2R \arcsin\sqrt{\sin^2{-0.019024088846736} + \cos{0.630412925820352} \cdot \cos{0.59236474812688} \cdot \sin^2{-0.2768964858289)}\ }<math>

<math>=2R \arcsin\sqrt{-0.01902294134728269363^2 + 0.80778416634911741504 \cdot 0.82962270331649880514 \cdot -0.27337167328001932836^2\ }<math>

<math>=2R \arcsin\sqrt{0.00036187229750215750 + 0.05008215253824176633}<math>

<math>=2R \arcsin\sqrt{0.05044402483574392383}<math>

<math>=2R \arcsin 0.22459747290596108537<math>

Substituting R with 6367000 meters, we get:

<math>\approx 2 \cdot 6367000\ \mathrm{m} \cdot 0.226529935868622<math>

<math>\approx 2884632.2\ \mathrm{m}<math>

The distance is 2884.6 km or 1792 miles.

Quality of the approximation

The shape of the Earth more closely resembles a flattened spheroid with extreme values for the radius of curvature of 6336 km at the equator and 6399 km at the poles. Using a sphere with a radius of 6367 km results in an error of up to about 0.5%.

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Categories: Trigonometry | Metric geometry