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Flight paths

You will learn about:
calculating distances on a sphere.

If you ever flew in a plane, you probably realized that the plane took a curved path instead of a straight path (if
you were looking at a flat map). This is because on a sphere, the shortest distance between two points is along a
great circle. This is more obvious when you plot the straight and curved paths on a flat map (which uses the
Mercator projection) and on a sphere.

You can calculate the length of the curved path using the haversine formula. Between two points on the Earth
$(\phi_1, \lambda_1)$ and $(\phi_2, \lambda_2)$ where $\phi$ is latitude and $\lambda$ is longitude, the shortest
distance is
$$ 2R \arcsin \sqrt{ \sin^2 \left( \frac{\phi_2 - \phi_1}{2} \right) + \cos\phi_1 \cos\phi_2 \sin^2 \left( \frac{\lambda_2 -
\lambda_1}{2} \right)} $$
where $R$ = 6372.1 km is the radius of the Earth.

Input:
(Latitude, longitude) coordinates of two points. Latitudes go from -90 to +90. Longitudes go from -180 to +180. Remember
to convert your angles from degrees to radians before using the sine and cosine functions!

Output:
The distance between the points in kilometers (km).

Example

Input point 1 (latitude, longitude) : 46.283, 86.667
Input point 2 (latitude, longitude) : -48.877, -123.393
Output distance: 17760.054

Difficulty

Timesink

Function

haversine_distance(lat1, lon1, lat2, lon2)

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Notes

In this problem we use the haversine formula
which calculates the great-circle distance
assuming the Earth is a perfect sphere so it can be off by up to 0.5%.
Vincenty's formulae account for the oblate
spheroid shape of the Earth and are much more accurate, being off only by <0.5 mm.

Let us know what you think about this problem! Was it too hard? Difficult to understand? Also feel free to
discuss the problem, ask questions, and post cool stuff on Discourse. You should be able see a discussion
thread below. Would be nice if you don't post solutions in there but if you do then please organize and
document your code well so others can learn from it.