Geographical distance

Distance, ${\displaystyle D,\,\!}$ is calculated between two points, ${\displaystyle P_{1}\,\!}$ and ${\displaystyle P_{2}\,\!}$. The geographical coordinates of the two points, as (latitude, longitude) pairs, are ${\displaystyle (\phi _{1},\lambda _{1})\,\!}$ and ${\displaystyle (\phi _{2},\lambda _{2}),\,\!}$ respectively. Which of the two points is designated as ${\displaystyle P_{1}\,\!}$ is not important for the calculation of distance.

Latitude and longitude coordinates on maps are usually expressed in degrees. In the given forms of the formulae below, one or more values must be expressed in the specified units to obtain the correct result. Where geographic coordinates are used as the argument of a trigonometric function, the values may be expressed in any angular units compatible with the method used to determine the value of the trigonometric function. Many electronic calculators allow calculations of trigonometric functions in either degrees or radians. The calculator mode must be compatible with the units used for geometric coordinates.

Differences in latitude and longitude are labeled and calculated as follows:

${\displaystyle {\begin{aligned}\Delta \phi &=\phi _{2}-\phi _{1};\\\Delta \lambda &=\lambda _{2}-\lambda _{1}.\end{aligned}}\,\!}$

It is not important whether the result is positive or negative when used in the formulae below.

"Mean latitude" is labeled and calculated as follows:

${\displaystyle \phi _{m}={\frac {\phi _{1}+\phi _{2}}{2}}.\,\!}$

Colatitude is labeled and calculated as follows:

For latitudes expressed in radians:

${\displaystyle \theta ={\frac {\pi }{2}}-\phi$ ;\,\!}

For latitudes expressed in degrees:

${\displaystyle \theta =90^{\circ }-\phi .\,\!}$

Unless specified otherwise, the radius of the Earth for the calculations below is:

${\displaystyle R\,\!}$ = 6,371.009 kilometers = 3,958.761 statute miles = 3,440.069 nautical miles.

${\displaystyle D_{\,}\!}$ = Distance between the two points, as measured along the surface of the Earth and in the same units as the value used for radius unless specified otherwise.

Longitude has singularities at the Poles (longitude is undefined) and a discontinuity at the ±180° meridian. Also, planar projections of the circles of constant latitude are highly curved near the Poles. Hence, the above equations for delta latitude/longitude (${\displaystyle \Delta \phi \!}$, ${\displaystyle \Delta \lambda \!}$) and mean latitude (${\displaystyle \phi _{m}\!}$) may not give the expected answer for positions near the Poles or the ±180° meridian. Consider e.g. the value of ${\displaystyle \Delta \lambda \!}$ ("east displacement") when ${\displaystyle \lambda _{1}\!}$ and ${\displaystyle \lambda _{2}\!}$ are on either side of the ±180° meridian, or the value of ${\displaystyle \phi _{m}\!}$ ("mean latitude") for the two positions (${\displaystyle \phi _{1}\!}$=89°, ${\displaystyle \lambda _{1}\!}$=45°) and (${\displaystyle \phi _{2}\!}$=89°, ${\displaystyle \lambda _{2}\!}$=−135°).

If a calculation based on latitude/longitude should be valid for all Earth positions, it should be verified that the discontinuity and the Poles are handled correctly. Another solution is to use n-vector instead of latitude/longitude, since this representation does not have discontinuities or singularities.

This formula takes into account the variation in distance between meridians with latitude:

${\displaystyle D=R{\sqrt {(\Delta \phi )^{2}+(\cos(\phi _{m})\Delta \lambda )^{2}}},}$

where:

${\displaystyle \Delta \phi \,\!}$ and ${\displaystyle \Delta \lambda \,\!}$ are in radians;

${\displaystyle \phi _{m}\,\!}$ must be in units compatible with the method used for determining ${\displaystyle \cos(\phi _{m}).\,\!}$

To convert latitude or longitude to radians use

${\displaystyle 1^{\circ }=(\pi /180)\,\mathrm {radians} .}$

This approximation is very fast and produces fairly accurate result for small distances . Also, when ordering locations by distance, such as in a database query, it is faster to order by squared distance, eliminating the need for computing the square root.

The FCC prescribes the following formulae for distances not exceeding 475 kilometres (295 mi):[2]

${\displaystyle D={\sqrt {(K_{1}\Delta \phi )^{2}+(K_{2}\Delta \lambda )^{2}}},}$

where

${\displaystyle D\,\!}$ = Distance in kilometers;

${\displaystyle \Delta \phi \,\!}$ and ${\displaystyle \Delta \lambda \,\!}$ are in degrees;

${\displaystyle \phi _{m}\,\!}$ must be in units compatible with the method used for determining ${\displaystyle \cos(\phi _{m});\,\!}$

${\displaystyle {\begin{aligned}K_{1}&=111.13209-0.56605\cos(2\phi _{m})+0.00120\cos(4\phi _{m});\\K_{2}&=111.41513\cos(\phi _{m})-0.09455\cos(3\phi _{m})+0.00012\cos(5\phi _{m}).\end{aligned}}\,\!}$

Where ${\displaystyle K_{1}}$ and ${\displaystyle K_{2}}$ are in units of kilometers per arc degree. It may be interesting to note that:

${\displaystyle K_{1}=M{\frac {\pi }{180}}\,\!}$ = kilometers per arc degree of latitude difference;

${\displaystyle K_{2}=\cos(\phi _{m})N{\frac {\pi }{180}}\,\!}$ = kilometers per arc degree of longitude difference;

where ${\displaystyle M\,\!}$ and ${\displaystyle N\,\!}$ are the meridional and its perpendicular, or "normal", radii of curvature (the expressions in the FCC formula are derived from the binomial series expansion form of ${\displaystyle M\,\!}$ and ${\displaystyle N\,\!}$, set to the Clarke 1866 reference ellipsoid).

For a more computationally efficient implementation of the formula above, multiple applications of cosine can be replaced with a single application and use of recurrence relation for Chebyshev polynomials.

${\displaystyle D=R{\sqrt {\theta _{1}^{2}\;{\boldsymbol {+}}\;\theta _{2}^{2}\;\mathbf {-} \;2\theta _{1}\theta _{2}\cos(\Delta \lambda )}},}$

where the colatitude values are in radians. For a latitude measured in degrees, the colatitude in radians may be calculated as follows: ${\displaystyle \theta ={\frac {\pi }{180}}(90^{\circ }-\phi ).\,\!}$

A tunnel between points on Earth is defined by a line through three-dimensional space between the points of interest. The great circle chord length may be calculated as follows for the corresponding unit sphere:

${\displaystyle {\begin{aligned}&\Delta {X}=\cos(\phi _{2})\cos(\lambda _{2})-\cos(\phi _{1})\cos(\lambda _{1});\\&\Delta {Y}=\cos(\phi _{2})\sin(\lambda _{2})-\cos(\phi _{1})\sin(\lambda _{1});\\&\Delta {Z}=\sin(\phi _{2})-\sin(\phi _{1});\\&C_{h}={\sqrt {(\Delta {X})^{2}+(\Delta {Y})^{2}+(\Delta {Z})^{2}}}.\end{aligned}}}$

The tunnel distance between points on the surface of a spherical Earth is ${\displaystyle D=RC_{h}}$. For short distances (${\displaystyle D\ll R}$), this underestimates the great circle distance by ${\displaystyle D(D/R)^{2}/24}$.

Lambert's formulae[13] give accuracy on the order of 10 meters over thousands of kilometers. First convert the latitudes ${\displaystyle \scriptstyle \phi _{1}}$, ${\displaystyle \scriptstyle \phi _{2}}$ of the two points to reduced latitudes ${\displaystyle \scriptstyle \beta _{1}}$, ${\displaystyle \scriptstyle \beta _{2}}$

${\displaystyle \tan \beta =(1-f)\tan \phi ,}$

where ${\displaystyle f}$ is the flattening. Then calculate the central angle ${\displaystyle \sigma }$ in radians between two points ${\displaystyle (\beta _{1},\;\lambda _{1})}$ and ${\displaystyle (\beta _{2},\;\lambda _{2})}$ on a sphere using the Great-circle distance method (law of cosines or haversine formula), with longitudes ${\displaystyle \lambda _{1}\;}$ and ${\displaystyle \lambda _{2}\;}$ being the same on the sphere as on the spheroid.

${\displaystyle P={\frac {\beta _{1}+\beta _{2}}{2}}\qquad Q={\frac {\beta _{2}-\beta _{1}}{2}}}$

${\displaystyle X=(\sigma -\sin \sigma ){\frac {\sin ^{2}P\cos ^{2}Q}{\cos ^{2}{\frac {\sigma }{2}}}}\qquad \qquad Y=(\sigma +\sin \sigma ){\frac {\cos ^{2}P\sin ^{2}Q}{\sin ^{2}{\frac {\sigma }{2}}}}}$

${\textstyle \mathrm {distance} =a{\bigl (}\sigma -{\tfrac {f}{2}}(X+Y){\bigr )}}$

where ${\displaystyle a}$ is the equatorial radius of the chosen spheroid.

On the GRS 80 spheroid Lambert's formula is off by

0 North 0 West to 40 North 120 West, 12.6 meters

0N 0W to 40N 60W, 6.6 meters

40N 0W to 40N 60W, 0.85 meter

Bowring maps the points to a sphere of radius R′, with latitude and longitude represented as φ′ and λ′. Define

${\displaystyle A={\sqrt {1+e'^{2}\cos ^{4}\phi _{1}}},\quad B={\sqrt {1+e'^{2}\cos ^{2}\phi _{1}}},}$

where the second eccentricity squared is

${\displaystyle e'^{2}={\frac {a^{2}-b^{2}}{b^{2}}}={\frac {f(2-f)}{(1-f)^{2}}}.}$

The spherical radius is

${\displaystyle R'={\frac {\sqrt {1+e'^{2}}}{B^{2}}}a.}$

(The Gaussian curvature of the ellipsoid at φ₁ is 1/R′².) The spherical coordinates are given by

${\displaystyle {\begin{aligned}\tan \phi _{1}'&={\frac {\tan \phi }{B}},\\\Delta \phi '&={\frac {\Delta \phi }{B}}{\biggl [}1+{\frac {3e'^{2}}{4B^{2}}}(\Delta \phi )\sin(2\phi _{1}+{\tfrac {2}{3}}\Delta \phi ){\biggr ]},\\\Delta \lambda '&=A\Delta \lambda ,\end{aligned}}}$

where ${\displaystyle \Delta \phi =\phi _{2}-\phi _{1}}$, ${\displaystyle \Delta \phi '=\phi _{2}'-\phi _{1}'}$, ${\displaystyle \Delta \lambda =\lambda _{2}-\lambda _{1}}$, ${\displaystyle \Delta \lambda '=\lambda _{2}'-\lambda _{1}'}$. The resulting problem on the sphere may be solved using the techniques for great-circle navigation to give approximations for the spheroidal distance and bearing. Detailed formulas are given by Rapp,[11] §6.5 and Bowring.[12]