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.¶¶For the forward conversion from latitude and longitude, for the south pole case¶¶E = FE + rho * sin(lon – lonO)¶N = FN + rho * cos(lon – lonO)¶where¶t = tan(pi/4 + lat/2) / {[(1 + e sin(lat)) / (1 – e sin(lat))]^(e/2)}¶rho = 2*a*ko*t / {[(1+e)^(1+e) (1–e)^(1–e)]^0.5}¶¶For the north pole case, ¶rho and E are found as for the south pole case but¶t  = tan(pi/4 – lat/2) *  {[(1 + e sin(lat)) / (1 – e sin(lat))]^(e/2)}¶N = FN – rho * cos(lon – lonO)¶¶For the reverse conversion from easting and northing to latitude and longitude,¶lat = chi + (e^2/2 + 5e^4/24 + e^6/12 + 13e^8/360) sin(2 chi) ¶+ (7e^4/48 + 29e^6/240 + 811e^8/11520) sin(4 chi)¶+ (7e^6/120 +  81e^8/1120) sin(6 chi)  + (4279e^8/161280) sin(8 chi)¶¶where rho'  = [(E-FE)^2  + (N – FN)^2]^0.5¶t'   =rho' {[(1+e)^(1+e) * (1– e)^(1-e)]^0.5} / (2 a ko)¶and for the south pole case¶chi  = 2 atan(t' ) – pi/2 ¶but for the north pole case¶chi  =  pi/2 - 2 atan t'¶¶Then for for both north and south cases if E = FE, lon = lonO¶else for the south pole case¶lon = lonO + atan [(E – FE) / (N – FN)]¶and for the north pole case¶lon = lonO + atan [(E – FE) / –(N – FN)] = lonO + atan [(E – FE) / (FN – N)]