Stereographic projection and uneven scaling

[Tahmeed]

Lets assume we are mapping one face of earth. we place a plane touching the Earth at 0 lattitude and 0 longitude. Now we take the plane of projection. suppose that we expand the projection unevenly. The small projectional area of a certain lattitude and longitude is expanded by a factor which is the function of it's lattitude abd longitude.

Evidently we won't get a circular projection. But how do i find the shape/equation of the projection if i know the expansion factor as a function of lattitude and longitude?

 

[.Scott]

The first thing you need to know is your map projection: https://en.wikipedia.org/wiki/Map_projection
The projection you are describing is Azimuthal, but there are variants.
From the wiki article:

Some azimuthal projections are true perspective projections; that is, they can be constructed mechanically, projecting the surface of the Earth by extending lines from a https://en.wikipedia.org/w/index.php?title=Point_of_perspective&action=edit&redlink=1 (along an infinite line through the tangent point and the tangent point's antipode) onto the plane:

• The gnomonic projection displays great circles as straight lines. Can be constructed by using a point of perspective at the center of the Earth. r(d) = c tan d/R; so that even just a hemisphere is already infinite in extent.[24][25]
• The General Perspective projection can be constructed by using a point of perspective outside the earth. Photographs of Earth (such as those from the International Space Station) give this perspective.
• The orthographic projection maps each point on the Earth to the closest point on the plane. Can be constructed from a point of perspective an infinite distance from the tangent point; r(d) = c sin d/R.[26] Can display up to a hemisphere on a finite circle. Photographs of Earth from far enough away, such as the Moon, approximate this perspective.
• The stereographic projection, which is conformal, can be constructed by using the tangent point's antipode as the point of perspective. r(d) = c tan d/2R; the scale is c/(2R cos2 d/2R).[27] Can display nearly the entire sphere's surface on a finite circle. The sphere's full surface requires an infinite map.

Other azimuthal projections are not true perspective projections:

• Azimuthal equidistant: r(d) = cd; it is used by amateur radio operators to know the direction to point their antennas toward a point and see the distance to it. Distance from the tangent point on the map is proportional to surface distance on the Earth (;[28] for the case where the tangent point is the North Pole, see the flag of the United Nations)
• Lambert azimuthal equal-area. Distance from the tangent point on the map is proportional to straight-line distance through the earth: r(d) = c sin d/2R[29]
• https://en.wikipedia.org/w/index.php?title=Logarithmic_azimuthal_projection&action=edit&redlink=1 is constructed so that each point's distance from the center of the map is the logarithm of its distance from the tangent point on the Earth. r(d) = c ln d/d0); locations closer than at a distance equal to the constant d0 are not shown.[30][31]

 

[math771]

There are a few different ways you could approach this problem, but here is one possible solution:

First, let's define our coordinate system. We will use the standard latitude and longitude coordinates, with 0 latitude being the equator and 0 longitude being the Prime Meridian. For simplicity, let's also assume that the expansion factor at the equator (latitude 0) and the Prime Meridian (longitude 0) is equal to 1.

Now, let's consider a point with coordinates (lat, long). We want to find the corresponding point on the projected plane, which we will call (x, y). We can think of this as a transformation from spherical coordinates (lat, long) to Cartesian coordinates (x, y).

First, let's consider the expansion factor at this point. We'll call it f(lat, long). We know that at the equator and Prime Meridian, f(lat, long) = 1. We can also assume that f(lat, long) is a continuous function that varies smoothly with latitude and longitude.

Next, we need to consider the angle between the point (lat, long) and the equator. We'll call this angle θ. We can calculate θ using basic trigonometry: θ = π/2 - lat. Note that at the equator, θ = 0, and at the poles (lat = ±90), θ = π/2.

Now, we can use the expansion factor and θ to calculate the distance from the projected point to the origin (0, 0). This distance will be equal to the radius of the projected circle at that point. We'll call this distance r. We can calculate r using the formula r = f(lat, long) * cos(θ).

Finally, we can use r to calculate the coordinates (x, y) of the projected point. We can do this using the standard polar-to-Cartesian conversion formulas: x = r * cos(long) and y = r * sin(long).

Putting all of this together, we have our transformation from spherical coordinates (lat, long) to Cartesian coordinates (x, y):

x = f(lat, long) * cos(long) * cos(π/2 - lat)
y = f(lat, long) * sin(long) * cos(π/2 - lat)

This will give you the shape of your projection, as it maps points on the sphere to points