# EPSG:1102

## Lambert Conic Conformal (1SP variant B)

### Attributes

Data source: EPSG

Information source: EPSG guidance note #7-2, https://epsg.org

Revision date: 2021-01-13

### Formula

```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.

To derive the projected Easting and Northing coordinates of a point with geographical coordinates (lat,lon):

E = Ef + r sin(theta)
N = Nf + rf - r cos(theta)
where
mO = cos(latO)/(1 – e^2 sin^2(latO))^0.5 where latO is the latitude of natural origin
tO = tan(pi/4 – latO/2)/[(1 – e sin(latO))/(1 + e sin(latO))]^e/2
tf = tan(pi/4 – latF/2)/[(1 – e sin(latF))/(1 + e sin(latF))]^e/2
t = tan(pi/4 – lat/2)/[(1 – e sin(lat))/(1 + e sin(lat))]^e/2
n = sin(latO)
F = mO/(n tO^n)
rf = a F tf^n kO
r =  a F t^n kO
lonF = lonO
theta = n(lon – lonO)
As with other conics, a negative n and r result for projections centered in the Southern Hemisphere.

The reverse formulas to derive the latitude and longitude of a point from its Easting and Northing values are:

lat = pi/2 - 2arctan{t'[(1 - e sin(lat))/(1 + e sin(lat))]^(e/2)}
lon = theta'/n +lonO
where
n, F, rf and lonO are derived as for the forward calculation
r' = +/-[(E - Ef)^2 + {rf - (N - Nf)}^2]^0.5  taking the sign of n
t' = (r'/(a k0 F))^(1/n)
If n is positive, theta' = atan2{(E –  Ef) , [rf – (N –  Nf)]}
but if n is negative the signs of both arguments of the atan2 function  must be reversed and theta' = atan2{– (E –  Ef) , – [rf – (N –  Nf)]}

Note that the formula for lat requires iteration. First calculate t' and then a trial value for lat using
lat = π/2-2atan(t'). Then use the full equation for lat substituting the trial value into the right hand side of the equation. Thus derive a new value for lat. Iterate the process until lat does not change significantly. The solution should quickly converge, in 3 or 4 iterations.```

### Example

```Parameters:
Ellipsoid:  GRS 1980, a = 6378137.000 m., 1/f = 298.2572221
then  e = 0.081819191 and e^2 = 0.006694380

Latitude of natural origin LatO      44°22'45"N  =  0.774562578 rad
Scale factor at natural origin ko   1.000000
Latitude of false origin LatF         45°11'00"N  =  0.788597934 rad
Longitude of false origin LongF      6°49'00"E  =  0.118973277 rad
Easting at false origin  Ef            150000.00 m
Northing at false origin Nf              50000.00 m

Forward calculation for:
first gives
mO    =  0.715900163        tO =  0.422551185           tf  =  0.414305398
n        =  0.699403505        F  =  1.869760448           rf  =  6439208.575
t         =  0.395846092       r    =  6237180.887     theta  =  0.002237931

Then Easting E   =     163958.366 m
Northing N =     252043.307 m

Reverse calculation for the same easting and northing first gives

r'        =  6237180.887
t'        =  0.395846092
theta' =  0.002237931

Then Latitude     = 47°00'00.000"N
Longitude   =  7°00'00.000"E```