# EPSG:1051

## Lambert Conic Conformal (2SP Michigan)

### Attributes

Data source: OGP

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

Revision date: 2018-08-29

#### 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 NAD27 / Michigan 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 = 44°11'00"N = 0.771144641 rad
Second Standard Parallel = 45°42'00"N = 0.797615468 rad
Latitude False Origin = 43°19'00"N = 0.756018454 rad
Longitude False Origin = 84°20'00"W = -1.471894336 rad
Easting at false origin = 2000000.00 US survey feet
Northing at false origin = 0.00 US survey feet
Ellipsoid scaling factor = 1.0000382

Forward calculation for:
Latitude = 43°45'00.00"N = 0.763581548 rad
Longitude = 83°10'00.00"W = -1.451532161 rad

first gives :
m1    = 0.718295175      m2 = 0.699629151
t        = 0.429057680      tF  = 0.433541026
t1      = 0.424588396      t2 = 0.409053868
n       = 0.706407410       F = 1.862317735
r        = 21436775.51    rF = 21594768.40
theta = 0.014383991

Then Easting X =      2308335.75 US survey feet
Northing Y =      160210.48 US survey feet

Reverse calculation for same easting and northing first gives:
theta' = 0.014383991     r' = 21436775.51
t'        = 0.429057680

Then Latitude     = 43°45'00.000"N
Longitude   = 83°10'00.000"W</epsg:example>
<gml:identifier codeSpace="IOGP">urn:ogc:def:method:EPSG::1051</gml:identifier>
<gml:name>Lambert Conic Conformal (2SP Michigan)</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 K F t^n         for rF and r, where rF is the radius of the parallel of latitude of the false origin and K is the ellipsoid scaling factor.
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 - 2atan{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'/(aKF))^(1/n)
theta' = atan2 [(E- EF),(rF - (N- NF))]   (see GN7-2 implementation notes in 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>