Editing Azimuthal equidistant projection (section)

== Mathematical definition ==
[[File:Azimuthal equidistant projection with Tissot's indicatrix.png | thumb | 220x124px | right | alt= Tissot's indicatrix applied to the azimuthal equidistant projection |
[[Tissot's indicatrix]] applied to the azimuthal equidistant projection]]
A point on the globe is chosen as "the center" in the sense that mapped distances and [[azimuth]] directions from that point to any other point will be correct. That point, (''φ''{{sub|0}}, ''λ''{{sub|0}}), will project to the center of a circular projection, with ''φ'' referring to latitude and ''λ'' referring to longitude. All points along a given azimuth will project along a straight line from the center, and the angle ''θ'' that the line subtends from the vertical is the azimuth angle. The distance from the center point to another projected point ''ρ'' is the [[arc length]] along a [[great circle]] between them on the globe. By this description, then, the point on the plane specified by (''θ'',''ρ'') will be projected to Cartesian coordinates:
:<math>x = \rho \sin \theta, \qquad y = -\rho \cos \theta</math>

The relationship between the coordinates (''θ'',''ρ'') of the point on the globe, and its latitude and longitude coordinates (''φ'', ''λ'') is given by the equations:
<ref name="Album">{{cite book
 |title        = An Album of Map Projections
 |last1        = Snyder
 |first1       = John P.
 |author-link1  = John P. Snyder
 |last2        = Voxland
 |first2       = Philip M.
 |author-link2  = Philip M. Voxland
 |year         = 1989
 |publisher    = [[United States Geological Survey|USGS]]
 |location     = Denver
 |series       = Professional Paper 1453
 |isbn         = 978-0160033681
 |pages        = 228
 |url          = https://pubs.er.usgs.gov/usgspubs/pp/pp1453
 |access-date   = 2018-03-29
 |archive-url  = https://web.archive.org/web/20100701102858/http://pubs.er.usgs.gov/usgspubs/pp/pp1453
 |archive-date = 2010-07-01
 |url-status   = dead
}}</ref>
:<math>
\begin{align}
\cos \frac{\rho}{R} &= \sin \varphi_0 \sin \varphi + \cos \varphi_0 \cos \varphi \cos \left(\lambda - \lambda_0\right) \\
\tan \theta &= \frac{\cos \varphi \sin \left(\lambda - \lambda_0\right)}{\cos \varphi_0 \sin \varphi - \sin \varphi_0 \cos \varphi \cos \left(\lambda - \lambda_0\right)}
\end{align}
</math>

When the center point is the north pole, ''φ''<sub>0</sub> equals <math>\pi/2</math> and ''λ''{{sub|0}} is arbitrary, so it is most convenient to assign it the value of 0. This assignment significantly simplifies the  equations for ''ρ''<sub>u</sub> and ''θ'' to:
:<math>\rho = R \left( \frac{\pi}{2} - \varphi \right), \qquad \theta = \lambda~~</math>