We can model the surface of the earth as a sphere of radius $R$. Even if in reality, the earth is more a kind of ellipsoid with an irregular surface, where each point of its surface owns an altitude in relation to the sea level.

Coordinates on earth with latitude and longitude
Coordinates on earth with latitude and longitude

Let $p1 = (\text{lat1}, \text{long1})$, $p2 = (\text{lat2}, \text{long2})$ be 2 points on the surface of the earth with a latitude and a longitude given in radian respectively. The latitude has a value between $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and the longitude between $]-\pi, \pi]$. So each point is identified by unique way with this tuple: latitude and longitude. There is a unique great circle passing both through $p1$ and $p2$. The smallest distance between $p1$ and $p2$ on the earth surface is the length of an arc circle of this great circle, that we note $d(p1, p2)$, the we put $\alpha \in [0, \pi]$ is the smallest center angle, to avoid all ambiguities, because $p1$ and $p2$ belongs to a circle.

Smallest distance between 2 points on the earth surface
Smallest distance between 2 points on the earth surface

So:

\[d(p1, p2) = R\;\alpha\]

With the formula of scalar product, we can find $\alpha$, because:

\[p1\,.\,p2 = ||p1||\,||p2||\,\cos(\alpha) = R^2 \cos(\alpha)\]

But also, in the cartesian coordinates system, with $( 0, 0, 0 )$ the center of the earth, we have that a point $p = (\text{lat}, \text{long})$ on the surface of the earth has for cartesian coordinates:

\[p = \begin{pmatrix} R\cos(\text{lat})\cos(\text{long}) \\\ R\cos(\text{lat})\sin(\text{long}) \\\ R\sin(\text{lat}) \end{pmatrix}\]

Indeed, even if it means changing reference points, we can always by symmetry come back to this diagram:

Cartesian coordinates on earth
Cartesian coordinates on earth

Then, we have also:

\[\begin{align*} p1\,.\,p2 &= R^2\left(\sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2})(\cos(\text{long1})\cos(\text{long2}) + \sin(\text{long1})\sin(\text{long2})\right) \\\ &= R^2\left(\sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2})\cos(\text{long1} - \text{long2})\right) \end{align*}\]

So:

\[\cos(\alpha) = \sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2})\cos(\text{long1} - \text{long2})\]

By applying the function $\text{arccos}: [-1, 1] \rightarrow [0, \pi]$ the reciprocal bijection of $\cos$ on $[0, \pi]$, we get the smallest center angle $\alpha$, and:

\[\boxed{ d(p1, p2) = R\,\text{arccos}\big(\sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2})\cos(\text{long1} - \text{long2})\big) }\]

For a historical reason and numerical computation often the haversine fonction $\text{hav}: \mathbb{R} \rightarrow [0, 1]\quad x \mapsto \sin^2\left(\frac{x}{2}\right)$.

We have $\cos(\alpha) = 1 - 2\text{hav}(\alpha)$, so:

\[1 - 2\text{hav}(\alpha) = \sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2}) - 2\text{hav}(\text{long1} - \text{long2})\cos(\text{lat1})\cos(\text{lat2})\]

That is to say:

\[\begin{align*} \text{hav}(\alpha) &= \dfrac{1 - (\sin(\text{lat1})\sin(\text{lat2}) + \cos(\text{lat1})\cos(\text{lat2}))}{2} + \text{hav}(\text{long1} - \text{long2})\cos(\text{lat1})\cos(\text{lat2}) \\\ &= \dfrac{1 - \cos(\text{lat1 - lat2})}{2} + \text{hav}(\text{long1} - \text{long2})\cos(\text{lat1})\cos(\text{lat2}) \end{align*}\]

Thus, we get the harvesine relation:

\[\boxed{ h(p1, p2) = \text{hav}(\text{lat1} - \text{lat2}) + \text{hav}(\text{long1} - \text{long2})\cos(\text{lat1})\cos(\text{lat2}) }\]

By using $\text{archav}: [0, 1] \rightarrow [0, \pi]$ the reciprocal bijection of $\text{hav}$ on $[0, \pi]$, we have for $\alpha \in [0, \pi]$:

\[d(p1, p2) = R\,\text{archav}(\text{hav}(\alpha))\]

In addition $\forall x \in [0, 1]$:

\[\text{archav}(x) = 2\,\text{arcsin}(\sqrt{x})\]

And by putting $y = \text{arcsin}(x)$ and $x = \sin(y)$ with $\tan(y) = \sqrt{\dfrac{\sin^2(y)}{1 - \sin^2(y)}}$ we have for $x \neq 1$:

\[\text{arcsin}(x) = \text{arctan}\left(\sqrt{\dfrac{x^2}{1 - x^2}}\right)\]

So:

\[d(p1, p2) = 2\,R\,\text{arctan}\left(\sqrt{\dfrac{\text{hav}(p1, p2)}{1 - \text{hav}(p1, p2)}}\right)\]

In order to have a valid formula for the case where $\text{hav}(p1, p2) = \text{hav}(\alpha) = 1$, that is to say for $\alpha = \pi$, we introduce the function:

\[\text{atan2}: \mathbb{R}^+ \times \mathbb{R}^+ \backslash \left\{0, 0\right\} \quad (x,y) \mapsto \left\{ \begin{array}{ll} \text{arctan}\left(\frac{x}{y}\right) & \text{if } y > 0 \\\ \frac{\pi}{2} & \text{if } y = 0 \end{array} \right.\]

So:

\[\boxed{ d(p1,p2) = 2\,R\,\text{atan2}\left( \sqrt{\text{hav}(p1, p2)}, \sqrt{1 - \text{hav}(p1, p2)} \right) }\]

We can check that the coherence of the formula where $d$ is well a metric distance, it is a positive function, symetric $d(p1, p2) = d(p2, p1)$, and homogeneous $d(p, p) = 0$ respecting the triangle inegality.

In practice, the radius of the earth is $R = 6371\,km$, and the latitude and longitude are given often in decimal degree and not in radian.


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