EPSG
US Geological Survey Professional Paper 1395; "Map Projections - A Working Manual"; J. Snyder.
2023-06-29
This is the ellipsoidal form of the projection.
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 equidistant conic projection is neither conformal nor equal area, but a compromise between the two. It is most frequently used for very small scale maps in atlases based on a sphere. EPSG carries only the ellipsoidal development in which the meridional arc distance between the parallels of latitude reflects that on the ellipsoid. The formulas to convert geodetic latitude and longitude (φ, λ) to easting (E) and northing (N) are: Easting, E = EF + r sin θ Northing, N = NF + rF – r cos θ where M = a [(1 – e^2/4 – 3e^4/64 – 5e^6/256 - …)φ – (3e^2/8 + 3e^4/32 + 45e^6/1024 + …)sin 2φ + (15e^4/256 + 45e^6/1024 + …)sin 4φ – (35e^6/3072 + …)sin 6φ + …] with φ in radians, and with MF, M1 and M2 for φF (latitude of false origin), φ1 (latitude of 1st standard parallel) and φ2 (latitude of 2nd standard parallel) derived in the same way; m = cos φ / (1 – e^2 sin2 φ)^0.5 with φ in radians, and m1 and m2 for φ1 and φ2 derived in the same way; n = a (m1 – m2) / (M2 – M1) G = (m1 / n) + (M1 / a) rF = a G – MF r = a G – M θ = n (λ – λF) For the reverse calculation: G and rF are constants for the projection calculated as for the forward calculation above. Then: r' = ± {(E – EF)^2 + [rF – (N – NF)]^2 }^0.5, taking the sign of n If n is positive (northern hemisphere case of the projection), θ' = atan2 {(E – EF) , [rF – (N – NF)]} but if n is negative (southern hemisphere case of the projection), the signs of both arguments of the atan2 function must be reversed: θ' = atan2 {– (E – EF), – [rF – (N – NF)]} M = a G – r' μ = M / [a (1 – e^2/4 – 3e^4/64 – 5e^6/256 - …)] Latitude, φ = μ + (3e1/2 – 27e1^3/32 + …)sin 2μ + (21e1^2/16 – 55e1^4/32 + …)sin 4μ + (151e1^3/96 - …)sin 6μ + (1097e1^4/512 - …)sin 8μ with μ in radians where e1 = [1 – (1 – e^2)^0.5] / [1 + (1 – e^2)^0.5] Longitude, λ = λF + (θ' / n)
Example (Northern Hemisphere): Parameters: Ellipsoid: Clarke 1866 a = 6378206.4 metres and b = 6356583.8 metres from which: 1/f =294.9786982, e = 0.08227185, e2 =0.00676866 and e1 = 0.001697916 Latitude of false origin = 23°00'00"N Longitude of false origin = 96°00'00"W Latitude of 1st standard parallel = 29°30'00"N Latitude of 2nd standard parallel = 45°30'00"N Easting at false origin = 0.00 metre Northing at false origin = 0.00 metre Forward calculation for: Latitude = 35°00'00.00"N Longitude = 75°00'00.00"W first gives : φF = 0.401425728 rad λF = -1.675516082 rad φ1 = 0.514872129 rad φ2 = 0.794124810 rad M1 = 3264511.20 m m1 = 0.871070821 M2 = 5040295.01 m m2 = 0.702119143 MF = 2544389.75 m n = 0.606835507 G = 1.947254290 rF = 9875600.03 m Then : φ = 0.610865238 rad λ = -1.308996939 rad M = 3874395.26 m m = 0.820065623 r = 8545594.52 m θ = 0.222416830 rad Then Easting = 1885051.86 m Northing = 1540507.64 m Reverse calculation for same easting and northing gives: r' = 8545594.52 m θ' = 0.222416830 rad M = 3874395.26 μ = 0.608473702 Then Latitude = 35°00'00.000"N Longitude = 75°00'00.000"W