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.¶¶Oblique aspect¶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 and Northing are:¶¶Easting, E   = FE + {(B . D) . [cos ß . sin(lon – lonO)]}¶Northing, N = FN + (B / D) . {(cos ßO . sin ß) –  [sin ßO . cos ß . cos(lon – lonO)]}¶¶where¶B = Rq . (2 / {1 + sin ßO . sin ß + [cos ßO . cos ß . cos(lon – lonO)]})^0.5¶D = a . [cos latO / (1 – e2 sin2 latO)^0.5] / (Rq . cos ßO)¶Rq = a . (qP  / 2)^0.5¶ß = asin (q / qP)¶ßO = asin (qO / qP)¶q = (1 – e^2) . ([sin(lat) / (1 – e^2 sin^2(lat))] – {[1/(2e)] . ln [(1 – e sin(lat)) / (1 + e sin(lat))]})¶qO = (1 – e^2) . ([sin(latO) / (1 – e^2 sin^2(latO))] – {[1/(2e)] . ln [(1 – e sin(latO)) / (1 + e sin(latO))]})¶qP = (1 – e^2) . ([sin(latP) / (1 – e^2 sin^2(latP))] – {[1/(2e)] . ln [(1 – e sin(latP)) / (1 + e sin(latP))]})¶where *P = p/2 radians, thus¶qP = (1 – e^2) . ([1 / (1 – e^2)] – {[1/(2e)] . ln [(1 – e) / (1 + e)]})¶¶The reverse formulas to derive the geodetic latitude and longitude of a point from its Easting and Northing values are:¶¶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 + atan {(E-FE) . sin C / [D. rho . cos ßO . cos C – D^2. (N-FN) . sin ßO . sin C]}¶where¶ß' = asin{(cosC . sin ßO) + [(D . (N-FN) . sinC . cos ßO) / rho]}¶C = 2 . asin(rho / 2 . Rq)¶rho = {[(E-FE)/D]^2 + [D . (N –FN)]^2}^0.5¶¶and D, Rq, and ßO are as in the forward equations.¶¶Polar aspect¶For the polar aspect of the Lambert Azimuthal Equal Area projection, some of the above equations are indeterminate. Instead, for the forward case from latitude and longitude (lat, lon) to Easting (E) and Northing (N):¶¶For the north polar case:¶	Easting, E   = FE + [rho  sin(lon – lonO)]¶	Northing, N = FN –  [rho  cos(lon – lonO)]¶where¶rho = a (qP  – q)^0.5¶and qP  and q are found as for the general case above.¶¶For the south polar case:¶	Easting, E   = FE + [rho . sin(lon – lonO)]¶	Northing, N = FN +  [rho . cos(lon – lonO)]¶where¶rho = a (qP  + q)^0.5¶and qP  and q are found as for the general case above.¶¶For the reverse formulas to derive the geodetic latitude and longitude of a point from its Easting and Northing:¶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']¶as for the oblique case, but where¶ß' = ±asin [1– rho^2 / (a^2{1– [(1– e^2)/2e)) ln[(1-e)/(1+ e)]})], taking the sign of  latO¶and rho = {[(E –FE)]^2 + [(N – FN)]^2}^0.5¶Then¶lon = lonO + atan [(E –FE)] / (N –FN)] for the south pole case¶and¶lon = lonO + atan [(E –FE)] / – (N –FN)] for the north pole case.