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 formulas to convert geodetic latitude and longitude (lat, lon) to Easting and Northing are:¶¶E = (rho . sin T) + FE¶N = (a . mO / sin(latO) –  rho . cos T) + FN¶¶where¶m = cos(lat) / (1 –  e^2sin^2(lat))^0.5¶with lat in radians and mO for latO, the latitude of the origin, derived in the same way.¶¶M = a[(1 –  e^2/4 –  3e^4/64 –  5e^6/256 –....)lat – (3e^2/8 + 3e^4/32 + 45e^6/1024+....)sin(2 lat) + (15e^4/256 + 45e^6/1024 +.....)sin(4 lat) –  (35e^6/3072 + ....)sin(6 lat) + .....]¶with lat in radians and MO for latO, the latitude of the origin, derived in the same way.¶¶rho = a . mO / sin(latO) + MO – M¶T = a . m (lon – lonO) / rho       with lon and lonO in radians¶¶For the reverse calculation:¶X = E – FE¶Y = N – FN¶rho = ± [X^2 + (a . mO / sin(latO) – Y)^2]^0.5  taking the sign of latO¶M = a . mO / sin(latO) + MO – rho¶mu = M / [a (1 – e^2/4 – 3e^4/64 – 5e^6/256 – …)]¶e1 = [1 – (1 – e^2)^0.5] / [1 + (1 – e^2)^0.5]¶lat = mu + ((3 e1 / 2) –  (27 e1^3 / 32) +.....)sin(2 mu) + ((21 e1^2 / 16) –  (55 e1^4 / 32) + ....)sin(4 mu) ¶  + ((151 e1^3 / 96) +.....)sin(6 mu) + ((1097 e1^4 / 512)  –  ....)sin(8 mu) + ......¶¶m = cos(lat) / (1 – e^2 sin^2(lat))^0.5¶¶If latO is not negative¶lon = lonO + rho {atan[X / (a . mO / sin(latO) – Y)]} / a . m¶but if lonO is negative¶lon = lonO + rho {atan[– X / (Y – a . mO / sin(latO))]} / a . m¶In either case, if lat = ±90°, m = 0 and the equation for lon is indeterminate, so use lon = lonO.