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.¶¶First calculate the scale factor at natural origin:¶for the south pole case¶tF  = tan (pi/4 + latF/2) / {[(1 + e sin(latF)) / (1 – e sin(latF))]^(e/2)}¶¶but for the north pole case¶tF  = tan (pi/4 - latF/2) * {[(1 + e sin(latF)) / (1 – e sin(latF))]^(e/2)}¶¶then for both cases¶mF = cos(latF)  / (1 – e^2 sin^2(latF))^0.5¶ko = mF {[(1+e)^(1+e) (1–e)^(1–e)]0.5} / (2  tF)¶¶
The forward and reverse conversions then follow the formulae for the¶Polar Stereographic (variant A) method:¶¶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)]