EPSG:9831

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 forward conversion from latitude and longitude is given by: 
	x = (lon – lonO) cos(lat) / [(1 – e^2 sin^2(lat))^0.5]
	E = FE + x
	N = FN + M – MO + ^2 tan(lat) [(1 – e^2 sin^2(lat))^0.5] / (2a)}
where
	M = a[(1 – ^2/4 –  3e^4/64 –  5e^6/256 –....)lat  –  (3e^2/8 + 3e^4/32 + 45e^6/1024+....)sin(2 lat) 
		+ (^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 natural origin, derived in the same way.

The reverse conversion from easting and northing to latitude and longitude requires iteration of three equations. The Guam projection uses three iterations, which is satisfactory over the small area of application. First MO for the latitude of the origin latO is derived as for the forward conversion. Then:
e'   = [1 – (1 –^2)^0.5] / [1 + (1 – e^2)^0.5]
and
M'  =  MO + (N – FN) – {(E – FE)^2 tan(latO) [(1 – e^2 sin^2(latO)^0.5] / (2a)}
mu'   =  M' / a(1 –  e^2/4 –  3e^4/64 –  5e^6/256 –....)
lat'   =  mu' + (3e'/2 –  27e'^3/32)sin(2mu') + (21e'^2/16 –  55e'^4/32)sin(4mu') + (151e'^3/96)sin(6mu') 
+ (1097e'^4/512)sin(8mu')

	M"  = MO + (N – FN) – {(E FE)^2 tan(lat') [(1 – e^2 sin^2(lat'))^0] / (2a)}
mu"   =  M" / a(1 –  e^2/4 –  3e^4/64 –  5e^6/256 –....)
lat"  =  mu" + (3e'/2 –  27e'^3/32)sin(2mu") + (21e'^2/16 –  55e'^4/32)sin(4mu") + (151e'^3/96)sin(6mu") 
	(1097  e'^4/512)sin(8mu")

	M'''   = MO + (N – FN) – {(E – F)^2 ta(lat") [(1 – e^2 sin^2(lat")^0.5] / (2a)}
mu'''   =  M''' / a(1 –  e^2/4 –  3e^4/64 –  5e^6/256 –....)
lat'''   =  mu''' + (3e'/2 –  27e'^3/32)sin(2mu''') + (21e'^2/16 –  55e'^4/32)sin(4mu''') + (151e'^3/96)sin(6mu''') 
+ (1097e'^4/512)sin(8mu''')
Then
lon = lonO + {(E – FE) . [(1 – e^2 sin^2 lat''')^0.5] / (a cos lat''')}