# Converting an Equitorial-Centered Azimuthal Projection into Equirectangular

The Problem

I made a map of a tidally-locked world under the premise that I was drawing it as an Azimuthal Projection, or a Planar Projection or something like that (I'm afraid I don't know the difference). This map is centered on the substellar point, where the star shines directly. I've been experimenting for some time on how to redraw this map as an equirectangular projection so that I can wrap it around a sphere in a program such as Google Earth, so that I can more easily visualize things like climates and winds. I have tried redrawing it by eye on a square half of a 1:2 rectangle, but I have not been able to master the way in which the circular gradations of distance from the substellar point (imagine them as circles of latitude, but originating from a point on the equator instead of the pole) slowly warp into a square as they approach the edges.

I've been using GIMP for this, and although I have searched the program for a tool to morph a circular image into a square one, I have not found one. Even if I had, this wouldn't be enough, since things get weird at the poles of equirectangular projections.

I have next to no experience in coding, and I don't know anything about how map projections work mathematically, and so any tips about programs I can use, methods I can employ, or resources I can consult are greatly appreciated.

• I'm pretty certain there are a bunch of past questions/answers on here that cover reprojecting maps, and appropriate tools for doing so. I don't have time to rummage through the archives right now, but you'd do well to dig around a bit. Oct 30, 2021 at 8:37
• i think Azimuthal Projections and Planar Projections are the same thing (from a quick google search)
– Nyra
Oct 31, 2021 at 6:34
• map projections are basically converting one set of points to another, for example if you take the latitude and longitude (which are angles that give points on a sphere) and instead treat them as $x$, $y$ type of values then you can map the points on the sphere to a flat plane. Hopefully that makes some sense
– Nyra
Oct 31, 2021 at 6:41
• this isn't something i play with myself but when i (briefly) considered making a realistic map for a project, someone recommended G.Projector (giss.nasa.gov/tools/gprojector/help/projections) which may or may not be helpful Nov 1, 2021 at 4:18
• Azmuithal is a particular type of planar projection. Planar just means projecting onto a flat surface (sheet of paper). There are loads of other planar projections. Mar 30, 2022 at 13:15

While I'm not an expert on maps, the projection you choose makes it easier converting the map to latitude an longitude.

Sorry in advance about the maths, I'll do my best to be as clear as i can (if it doesn't make sense please comment)

The azimuthal equidistant projection converts points on a sphere to a flat plane. It does this by taking the the latitude and the longitude (which i'm going to use the letters $$\phi$$ an $$\theta$$ to represent). So $$\phi$$, the latitude, is between $$-90°$$ and $$90°$$, and is the angular distance from the equator; and $$\theta$$, the longitude, is between $$0°$$ and $$180°$$, and is the angular distance around the globe from an arbitrary point.

The azimuthal projection converts these points, ($$\phi,\theta$$), to a flat plane with rectangular points ($$x,y$$) which are normal $$x$$ and $$y$$ points. It does this by "mapping" the arbitrary point on the sphere ($$\phi,\theta$$) to a point on the plane ($$x,y$$), by saying

$$x=\rho \cos(\theta)$$ $$y=-\rho \sin(\theta)$$

Where $$\rho$$ is the distance from a fixed point on the sphere to the arbitrary point. If the fixed point is at $$\phi=90°$$, then $$\rho=\pi R_{planet}\left(\frac{\phi+90°}{180°} \right)$$

where $$R_{planet}$$ is the radius of the planet.

So the relation ship between the latitude and the longitude, ($$\phi,\theta$$), and the points on the azimuthal projection, ($$x,y$$), are

$$x=\pi R_{planet}\left(\frac{\phi+90°}{180°} \right) \cos(\theta)$$ $$y=-\pi R_{planet}\left(\frac{\phi+90°}{180°} \right) \sin(\theta)$$

and to convert back

$$\phi=\frac{180}{R_{planet}\pi}\sqrt{x^2+y^2}+90°$$ $$\theta=-\arctan(\frac{y}{x}$$

A possible method of implementing the is drawing regions, of equal steps in latitude and longitude, on you map. Then seeing where the regions map to, on a square projection, such as the Mercator projection.

Don't know how GIMP works so I'm afraid I can't offer advice on how to implement this more accurately.

hopefully that helps

• (1) There is no such thing as "the" azimuthal projection. The specific projection you are discussing is called azimuthal equidistant. (A map projection is called azimuthal if it preserves the directions, or azimuths, from a fixed point to any other point. The azimuthal perspective and azimuthal equal-area projections are widely used, for example.) (2) The question explicitly states that the map is centered on the substellar point, whereas your formulae assume that projection is centerd on the north pole. Oct 31, 2021 at 10:21
• 1) corrected, thanks, I am more familiar with mathematical transformations than map projections 2) While latitude and longitude when used on earth have the north pole at latitude = 90°, the coordinates are arbitrary and it seemed a logical choice of coordinates to set latitude = 90° to be the subsolar point as there is rotation of the sun around the planet.
– Nyra
Oct 31, 2021 at 10:40