Converting an Equitorial-Centered Azimuthal Projection into Equirectangular

Question

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.

Answer 1

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