# EPSG:9803

## Lambert Conic Conformal (2SP Belgium)

### Attributes

Data source: OGP

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

Revision date: 2018-08-29

Remarks: In 2000 this modification was replaced through use of the regular Lambert Conic Conformal (2SP) method  with appropriately modified parameter values.

#### Export

Definition: OGP XML

```<?xml version="1.0" encoding="UTF-8"?>
<epsg:informationSource>EPSG guidance note #7-2, http://www.epsg.org</epsg:informationSource>
<epsg:revisionDate>2018-08-29</epsg:revisionDate>
<epsg:changes>
</epsg:changes>
<epsg:show>true</epsg:show>
<epsg:isDeprecated>false</epsg:isDeprecated>
<epsg:isOperationReversible>true</epsg:isOperationReversible>
<epsg:example>For Projected Coordinate System Belge 1972 / Belge Lambert 72

Parameters:
Ellipsoid  International 1924,  a = 6378388 metres
1/f = 297
then e = 0.08199189 and e^2 = 0.006722670

First Standard Parallel        49°50'00"N       =   0.86975574 rad
Second Standard Parallel  51°10'00"N       =   0.89302680 rad
Latitude False Origin          90°00'00"N       =   1.57079633 rad
Longitude False Origin         4°21'24.983"E = 0.07604294 rad
Easting at false origin EF        150000.01  metres
Northing at false origin NF    5400088.44  metres

Forward calculation for:

first gives :
m1     = 0.64628304         m2 = 0.62834001
t        = 0.59686306          tF  = 0.00000000
t1      = 0.36750382           t2 = 0.35433583
n       = 0.77164219            F = 1.81329763
r        = 37565039.86         rF = 0.00
alpha = 0.00014204     theta = 0.01953396

Then Easting E  =      251763.20 metres
Northing N =      153034.13 metres

Reverse calculation for same easting and northing first gives:
theta' = 0.01939192      r' = 548041.03
t' = 0.35913403
Then Latitude   =    50°40'46.461"N
Longitude =     5°48'26.533"E</epsg:example>
<gml:identifier codeSpace="IOGP">urn:ogc:def:method:EPSG::9803</gml:identifier>
<gml:name>Lambert Conic Conformal (2SP Belgium)</gml:name>
<gml:remarks>In 2000 this modification was replaced through use of the regular Lambert Conic Conformal (2SP) method  with appropriately modified parameter values.</gml:remarks>
<gml: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.

For the Lambert Conic Conformal (2 SP Belgium), the formulas for the regular two standard parallel case (coordinate operation method code 9802) are used except for easting, northing in the forward formula and lon in the rverse formula.

To derive the projected Easting and Northing coordinates of a point with geographical coordinates (lat,lon) the formulas for the one standard parallel case are:

Easting, E = EF + r sin (theta - alpha)
Northing, N = NF + rF - r cos (theta - alpha)
where
m = cos(lat)/(1 - e^2 sin^2(lat))^0.5     for m1, lat1, and m2, lat2 where lat1 and lat2 are the latitudes of the two standard parallels.
t  = tan(pi/4 - lat/2)/[(1 - e sin(lat))/(1 + e sin(lat))]^(e/2)   for t1, t2, tF and t using lat1, lat2, latF and lat respectively.
n = (loge(m1) - loge(m2))/(loge(t1) - loge(t2))
F = m1/(n  t1^n)
r =  a F t^n         for rF and r, where rF is the radius of the parallel of latitude of the false origin.
theta = n(lon - lon0)
alpha = 29.2985 seconds.

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 - esin(lat))/(1 + esin(lat))]^(e/2)}
lon = ((theta' + alpha)/n) +lon0
where
r' = +/-[(E - EF)^2 + {rF - (N - NF)}^2]^0.5 , taking the sign of n
t' = (r'/(aF))^(1/n)
theta' = atan2 [(E- EF),(rF - (N- NF))]
(see implementation notes in GN7-2 preface for atan2 convention)
alpha = 29.2985 seconds
and n, F, and rF are derived as for the forward calculation.

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.</gml:formula>