Coordinate System – Understanding ‘Ographic’ Latitude in IAU Projection 39916

coordinate systempyproj

In pyproj, I defined/loaded projection IAU 39916. This has the name "Earth (2015) / Ographic / Equirectangular, clon = 180". (I was looking for an equirectangular projection for representing an area on both hemispheres around the antimeridian.)

What does "ographic" mean here? In this context, it cannot mean orthographic, because the projection is equirectangular and not orthographic. Google Search gives me results on converting between "ographic" and "ocentric" (example 1, 2), but I have not found what those actually are. Is it related to geocentric and geodetic latitudes in any way?

In [292]: from pyproj import CRS

In [293]: crs = CRS.from_authority("IAU_2015", 39916)

In [294]: print(crs.name)
Earth (2015) / Ographic / Equirectangular, clon = 180

Best Answer

Ocentric and ographic latitudes do not refer to a positive East vs. positive West distinction.

Ocentric and ographic instead refer to where one measures latitude from. Direct from the proj docs:

The geodetic (or geographic) latitude (also called planetographic latitude in the context of non-Earth bodies) is the angle between the equatorial plane and the normal (vertical) to the ellipsoid surface at the considered point. The geodetic latitude is what is normally used everywhere in PROJ when angular coordinates are expected or produced.

The geocentric latitude (also called planetocentric latitude in the context of non-Earth bodies) is the angle between the equatorial plane and a line joining the body centre to the considered point.

https://proj.org/operations/conversions/geoc.html

The only time that these values will differ is when the body is non-spherical. In the planetary case, the Moon is defined as a sphere and ographic lat = ocentric lat. For non-spherical definitions of Mars ocentric lat != ographic lat.

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[GIS] Mercator projection: problem with latitude formula

The formula and its application are basically correct: the length of the Equator is 2 * pi * R and equating that with 707 pixels tells us the scale of the map at the Equator. However, any formula involving trigonometric functions of angles that involves pi expects the angles to be given in radians. Therefore, the Mercator position of latitude 51 degrees must be computed as

707 / (2 * pi) * ln(tan(pi / 4 Radians + 51 / 2 Degrees))
= 707 / (2 * pi) * ln(tan(pi / 4 + (51/2) * pi / 180))
= 707 / (2 * pi) * ln(tan(1.23046))
= 116.8

pixels above the Equator.

If degrees are treated as if they were radians, then pi/4 + 51/2 radians is the same as 42.0848 degrees (plus four whole circles--8*360--which makes no difference), which indeed places the location in Spain rather than Belgium. In this case it seems like a relatively small error of just 117 - 91 = 26 pixels has occurred, which can be misleading, because at other latitudes the result would be incredibly far off or undefined. For instance, using 52 instead of 51 in the formula gives a negative tangent whose logarithm is undefined. Increasing to 54.977871437821385 places you 3786 pixels below the Equator!

[GIS] a radar projection

The radar projection is an azimuthal projection given as this x = R sin (azimuth) y = R cos (azimuth) (Azimuth increasing clockwise from north, and R is range). On the map, straight radial lines from the radar are along a constant radar azimuth. The exact type of azimuthal projection depends on what "R" is.

If R is a great circle distance (i.e. to the point on the surface under the radar beam) then this is an oblique azimuthal equidistant projection.
If the distance R is along the radar beam itself then things become complicated in because the answer depends on the elevation angle of the antenna and how much the atmosphere will bend the beam. To a first approximation, if the radar's antenna elevation angle is 0 (horizontal) then R along the beam makes the projection equivalent to an gnomonic projection with the Earth's radius, Re, replaced by 4/3 Re.

So unfortunately you need to know what R was used in the map (grid) creation. In practice if one chooses any azimuthal projection the error will be negligible is within about 100km of a radar and for most purposes you will probably be OK to 200km.

Best Answer

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[GIS] Mercator projection: problem with latitude formula

[GIS] a radar projection

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