EPSG:9802

Lambert Conic Conformal (2SP)

Attributes

Data source: OGP

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

Revision date: 2018-08-29

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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 NAD27 / Texas South Central

Parameters:
Ellipsoid  Clarke 1866, a = 6378206.400 metres = 20925832.16 US survey feet
1/f = 294.97870
then e = 0.08227185 and e^2 = 0.00676866

First Standard Parallel          28°23'00"N  =   0.49538262 rad
Second Standard Parallel    30°17'00"N  =   0.52854388 rad
Latitude False Origin            27°50'00"N  =   0.48578331 rad
Longitude False Origin         99°00'00"W = -1.72787596 rad
Easting at false origin           2000000.00  US survey feet
Northing at false origin          0.00  US survey feet

Forward calculation for:

first gives :
m1    = 0.88046050      m2 = 0.86428642
t        = 0.59686306      tF  = 0.60475101
t1      = 0.59823957      t2 = 0.57602212
n       = 0.48991263       F = 2.31154807
r        = 37565039.86    rF = 37807441.20
theta = 0.02565177

Then Easting E =      2963503.91 US survey feet
Northing N =      254759.80 US survey feet

Reverse calculation for same easting and northing first gives:
theta' = 0.025651765     r' = 37565039.86
t'        = 0.59686306

Then Latitude     = 28°30'00.000"N
Longitude   = 96°00'00.000"W</epsg:example>
<gml:identifier codeSpace="IOGP">urn:ogc:def:method:EPSG::9802</gml:identifier>
<gml:name>Lambert Conic Conformal (2SP)</gml:name>
<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.

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:

E = EF + r sin(theta)
N = NF + rF - r cos(theta)
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)

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'/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)
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>