# EPSG:1078

## Equal Earth

### Attributes

Data source: EPSG

Information source: Šavrič et al, International Journal of Geographical Iinformation Science, August 2018.

Revision date: 2019-05-03

### Formula

```Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided.

The Equal Earth projection is an equal-area pseudocylindrical projection. It is appropriate for mapping global phenomena or for any other thematic world map that requires areas at their correct relative sizes. Its key features are its resemblance to the Robinson projection and continents with a visually pleasing appearance similar to those found on a globe.

Forward:
To derive the projected coordinates of a point, geodetic latitude (lat) is converted to authalic latitude (ß). The formulae to convert geodetic latitude and longitude (lat,lon) to Easting (E) and Northing (N) are:

Easting, E = FE + Rq 2 (lon – lonO) cos(θ) / {√3 [1.340264 – 0.243318 θ^2 + θ^6 (0.006251 + 0.034164 θ^2)]}
Northing, N = FN + Rq θ [1.340264 – 0.081106 θ^2 + θ^6 (0.000893 + 0.003796 θ^2)]

where
θ	=	asin [sin(ß) √3 / 2]
Rq	=	a (qP  / 2)^0.5
ß	=	asin(q / qP)
q	=	(1 – e^2) ({sin(lat)/ [1 – e^2 sin^2(lat)]} – 1/(2e) ln{[1 – e sin(lat)] / [1 + e sin(lat)]})

qP	=	(1 – e^2) ({sin(latP)/ [1 – e^2 sin^2(latP)]} – 1/(2e) ln{[1 – e sin(latP)] / [1 + e sin(latP)]})

where latP = π/2 radians, thus

qP	=	(1 – e^2) ([1 / (1 – e^2)] – {[1/(2e)]  ln [(1 – e) / (1 + e)]})

Reverse:
The reverse conversion from easting and northing to latitude and longitude requires iteration of northing (N) equation to obtain θ. θ0 = (N–FN) / Rq  is used as the first trial θ. A correction Δθn is calculated and subtracted from the trial θn value to obtain the next trial θn+1. n is the suffix number of iteration.

θ0 = (N–FN) / Rq
Δθ0 = { θ0 [1.340264 – 0.081106 θ0^2 + θ0^6 (0.000893 + 0.003796 θ0^2)] – [N–FN] / Rq} /
{1.340264 – 0.243318 θ0^2 + θ0^6 [0.006251 + 0.034164 θ0^2]}
θ1 = θ0 – Δθ0

Δθ1 = {θ1 [1.340264 – 0.081106 θ1^2 + θ1^6 (0.000893 + 0.003796 θ1^2)] – [N–FN] / Rq} /
{1.340264 – 0.243318 θ1^2 + θ1^6 [0.006251 + 0.034164 θ1^2]}
θ2 = θ1 – Δθ1

etc.

The calculation is repeated until Δθn is less than a predetermined convergence value. Then, using the final θn+1 as θ, the geodetic latitude and longitude of a point are determined as follows:

lat = ß + {[(e^2/3 + 31e^4/180 + 517e^6/5040) sin(2ß)] + [(23e^4/360 + 251e^6/3780) sin(4ß)]
+ [(761e^6/45360) sin(6ß)]}
lon = lonO + √3 (E–FE) {1.340264 – 0.243318 θ^2 + θ^6 (0.006251 + 0.034164 θ^2)} / {2 Rq cos(θ) }

where  ß = asin{ 2 sin(θ) / √3} and Rq is defined as in the forward equations.

Sphere:
For the spherical form of the projection, ß = lat  and  Rq = R, where R is the radius of the sphere.```

### Example

```Parameters:
Ellipsoid:	WGS 1984	a = 6378137.0 metres		1/f =298.257223563
then 	e = 0.08181919084262

Longitude of natural origin (lonO) = 90°00'00.00"W = -1.5707963268 rad
False easting (FE) = 0.000 m
False northing (FN) = 0.000 m

Forward calculation for:
Latitude = 34°03'27.169" N	=	 0.5944163293   rad
Longitude = 117°11'48.349" W	=	-2.0454693977   rad

First gives
ß =	0.5923399644	Rq =	6371007.181
θ =	0.5046548375

whence
E = -2390749.042 m
N =  4242849.758 m

Reverse calculation for the same Easting and Northing (-2390749.042 E, 4242849.758 N) starts with an iteration to obtain θ:

(N–FN) / Rq =	0.665962169
Step 1:	θ0 =	0.665962169	Δθ0 =	 0.164312383
Step 2:	θ1 =	0.501649786	Δθ1 =	-0.003004201
Step 3:	θ2 =	0.504653987	Δθ2 =	-8.512742e-07
Step 4:	θ3 =	0.504654838	Δθ3 =	-6.859963e-14
θ = θ4 =	0.504654838

This gives:
ß =	0.592339965

Then	Latitude =   34°03'27.169" N
Longitude = 117°11'48.349" W```