Azimuthal equidistant projection

While it may have been used by ancient Egyptians for star maps in some holy books,{{Cite book

| publisher = University of Chicago Press

| isbn = 0-226-76747-7

| last = SNYDER

| first = John P.

| title = Flattening the earth: two thousand years of map projections

| date = 1997

}}, p.29 the earliest text describing the azimuthal equidistant projection is an 11th-century work by al-Biruni.David A. KING (1996), "Astronomy and Islamic society: Qibla, gnomics and timekeeping", in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 1, p. 128–184 [153]. Routledge, London and New York. The first use of the projection on a terrestrial map is the pair of hemispheres by Glareanus of about 1510. Another early example of this system is the world map by ‛Ali b. Ahmad al-Sharafi of Sfax in 1571. Edward S. Kennedy, 1996, Mathematical geography, in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 1, Routledge, London and New York.

The projection appears in many Renaissance maps, and Gerardus Mercator used it for an inset of the north polar regions in sheet 13 and legend 6 of his well-known 1569 map. In France and Russia this projection is named "Postel projection" after Guillaume Postel, who used it for a map in 1581.Snyder 1997, p. 29 Many modern star chart planispheres use the polar azimuthal equidistant projection.

{{comparison_azimuthal_projections.svg}}

The polar azimuthal equidistant projection has also been adopted by 21st century Flat Earthers as a map of the Flat Earth, particularly due to its use in the UN flag and its depiction of Antarctica as a ring around the edge of the Earth.{{Cite web |date=2022-09-12 |title=Fact Check: The UN Flag Does NOT Show Flat Earth {{!}} Lead Stories |url=https://leadstories.com/hoax-alert/2022/09/fact-check-the-un-flag-does-not-show-flat-earth.html |access-date=2024-02-04 |website=leadstories.com |language=en-US}}

[[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:

:$x\; =\; \backslash rho\; \backslash sin\; \backslash theta,\; \backslash qquad\; y\; =\; -\backslash rho\; \backslash cos\; \backslash theta$

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:

{{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 = 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

}}

:$$

\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}

When the center point is the north pole, φ₀ equals $\backslash pi/2$ 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_u and θ to:

:$\backslash rho\; =\; R\; \backslash left(\; \backslash frac\{\backslash pi\}\{2\}\; -\; \backslash varphi\; \backslash right),\; \backslash qquad\; \backslash theta\; =\; \backslash lambda~~$

With the circumference of the Earth being approximately {{convert|40,000|km|mi|0|abbr=on}}, the maximum distance that can be displayed on an azimuthal equidistant projection map is half the circumference, or about {{convert|20,000|km|mi|0|abbr=on}}. For distances less than {{convert|10,000|km|mi|0|abbr=on}} distortions are minimal.{{efn|The country boundaries and landmasses in the azimuthal equidistant maps below look similar to the way they look on globes within the regions inside the red circles of radius {{convert|10,000|km|mi|0|abbr=on}}}} For distances {{convert|10,000|-|15,000|km|mi|0|abbr=on}} the distortions are moderate.{{efn|The Indian subcontinent and Arabian Peninsula are between the red and purple circles in the azimuthal equidistant map below centered about Los Angeles (circle radii between {{convert|10,000|-|15,000|km|mi|0|abbr=on}}), and hence look squished, yet still recognizable, relative to the way they look on a globe}} Distances greater than {{convert|15,000|km|mi|0|abbr=on}} are severely distorted.{{efn|The Island of Madagascar, off the East Coast of Africa is outside the purple circle of radius {{convert|15,000|km|mi|0|abbr=on}} in the azimuthal equidistant map below centered about Los Angeles, and hence looks much more stretched out than it would on a globe}}

If the azimuthal equidistant projection map is centered about a point whose antipodal point lies on land and the map is extended to the maximum distance of {{convert|20,000|km|mi|0|abbr=on}} the antipode point smears into a large circle. This is shown in the example of two maps centered about Los Angeles, and Taipei. The antipode for Los Angeles is in the south Indian Ocean hence there is not much significant distortion of land masses for the Los Angeles centered map except for East Africa and Madagascar. On the other hand, Taipei's antipode is near the Argentina–Paraguay border, causing the Taipei centered map to severely distort South America.

{{Multiple image

| direction = horizontal

| width = 260

| image1 = Los Angeles centered azimuthal equidistant projection.gif

| alt1 = An azimuthal equidistant projection centered about Los Angeles.

| caption1 = Map centered about Los Angeles, whose antipodal point is in the south Indian Ocean
{{hlist|{{legend0|#21B04B|California}}}}

| image2 = Taipei centered azimuthal equidistant projection.gif

| alt2 = An azimuthal equidistant projection centered about Taipei.

| caption2 = Map centered about Taipei, whose antipodal point is near the Argentina–Paraguay border
{{hlist|{{legend0|#21B04B|Taiwan}}|{{legend0|#FFFF00|Brazil}}|{{legend0|#FFA500|Paraguay}}|{{legend0|#FFC0CB|Argentina}}|{{legend0|#ADFF2F|Chile}}|{{legend0|#DDA0DD|Bolivia}}|{{legend0|#DCDCDC|Uruguay}}}}

| header = Sample azimuthal equidistant projection maps of maximum distance
(radius ≈ {{convert|20,000|km|mi|0|abbr=on}})

| footer = {{legend-line|red solid 2px|Red circles: 10,000 km radius circles}} {{legend-line|purple solid 2px|Purple circles: 15,000 km radius circles}}

| footer_align = center

| header_align = center

| align = center

}}

File:Gcmsyd.jpg

Azimuthal equidistant projection maps can be useful in terrestrial point to point communication. This type of projection allows the operator to easily determine in which direction to point their directional antenna. The operator simply finds on the map the location of the target transmitter or receiver (i.e. the other antenna being communicated with) and uses the map to determine the azimuth angle needed to point the operator's antenna. The operator would use an electric rotator to point the antenna. The map can also be used in one way communication. For example if the operator is looking to receive signals from a distant radio station, this type of projection could help identify the direction of the distant radio station. In order for the map to be useful, the map should be centered as close as possible about the location of the operator's antenna.{{Citation needed|date=October 2016}}

File:North Korean missile range.svg]]

Azimuthal equidistant projection maps can also be useful to show ranges of ballistic missiles, as demonstrated by the map centered on North Korea showing the country's missile range.

• Lambert azimuthal equal-area projection
• List of map projections
• Modern flat Earth beliefs

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{{Reflist}}

• [http://www.radicalcartography.net/?projectionref Table of examples and properties of all common projections], from radicalcartography.net
• [http://www.wm7d.net/azproj.shtml Online Azimuthal Equidistant Map Generator]
• [https://web.archive.org/web/20070927225345/http://www.uff.br/mapprojections/AzimuthalEquidistant_en.html An interactive Java Applet to study the metric deformations of the Azimuthal Equidistant Projection].
• [https://geographiclib.sourceforge.io GeographicLib] provides a class for performing azimuthal equidistant projections centered at any point on the ellipsoid.
• [http://earth.nullschool.net/#current/wind/isobaric/500hPa/azimuthal_equidistant=24.64,98.15,169 Animated US National Weather Service Wind Data for Azimuthal equidistant projection].
• [https://ns6t.net/azimuth/azimuth.html Generate an Azimuthal equidistant projection from any point on Earth] from ns6t.net.

{{Map projections}}

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Category:Equidistant projections